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1) Let abcd be a general four-digit number and all the digits are non-zero. How many four-digits numbers abcd exist such that the four digits are all distinct and such that a + b + c = d?
(A) 6
(B) 7
(C) 24
(D) 36
(E) 42
2) Let abcd be a general four-digit number. How many odd four-digits numbers abcd exist such that the four digits are all distinct, no digit is zero, and the product of a and b is the two digit number cd?
(A) 4
(B) 6
(C) 12
(D) 24
(E) 36
3) There are 500 cars on a sales lot, all of which have either two doors or four doors. There are 165 two-door cars on the lot. There are 120 four-door cars that have a back-up camera. Eighteen percent of all the cars with back-up cameras have standard transmission. If 40% of all the cars with both back-up cameras and standard transmission are two-door cars, how many four-door cars have both back-up cameras and standard transmission?
(A) 18
(B) 27
(C) 36
(D) 45
(E) 54
4) At Mnemosyne Middle School, there are 700 students: all the students are boys or girls in the 4^{th} or 5^{th} grade. There are 320 students in the 4^{th} grade, and there are 210 girls in the 5^{th} grade. Fifty percent of the 5^{th} graders and 40% of the 4^{th} graders take Mandarin Chinese. Ninety 5^{th} grade boys do not take Mandarin Chinese. The number of 4^{th} grade girls taking Mandarin Chinese is less than half of the number of 5^{th} grade girls taking Mandarin Chinese. Which of the following could be the number of 4^{th} grade boys in Mandarin Chinese?
(A) 10
(B) 40
(C) 70
(D) 100
(E) 130
5) A hundred identical cubic boxes are currently arranged in four cubes: a single cubic box, a 2 x 2 x 2 cube, a 3 x 3 x 3 cube, and a 4 x 4 x 4 cube. These four are not touching each other. All outward faces are painted and all inward faces are not painted. These four cubes are going to be dismantled and reassembled as a flat 10 x 10 square. The top and all the edges of this 10 x 10 square must be painted, but there is no requirement for paint on the bottom. How many individual faces will have to be painted to accommodate the requirements of this new design?
(A) 0
(B) 5
(C) 9
(D) 16
(E) 27
6) Twelve points are spaced evenly around a circle, lettered from A to L. Let N be the total number of isosceles triangles, including equilateral triangles, that can be constructed from three of these points. A different orientation of the same lengths counts as a different triangle, because a different combination of points form the vertices. What is the value of N?
(A) 48
(B) 52
(C) 60
(D) 72
(E) 120
7) Theresa is a basketball player practicing her free throws. On her first free throw, she has a 60% chance of making the basket. If she has just made a basket on her previous throw, she has a 80% of making the next basket. If she has just failed to make a basket on her previous throw, she has a 40% of making the next basket. What is the probability that, in five throws, she will make at least four baskets?
8) Suppose a “Secret Pair” number is a four-digit number in which two adjacent digits are equal and the other two digits are not equal to either one of that pair or each other. For example, 2209 and 1600 are “Secret Pair” numbers, but 1333 or 2552 are not. How many “Secret Pair” numbers are there?
(A) 720
(B) 1440
(C) 1800
(D) 1944
(E) 2160
9) In the coordinate plane, a circle with its center on the negative x-axis has a radius of 12 units, and passes through (0, 6) and (0, – 6). What is the area of the part of this circle in the first quadrant?
10) In the coordinate plane, line L passes above the points (50, 70) and (100, 89) but below the point (80, 84). Which of the following could be the slope of line L?
(A) 0
(B) 1/2
(C) 1/4
(D) 2/5
(E) 6/7
11) At the beginning of the year, an item had a price of A. At the end of January, the price was increased by 60%. At the end of February, the new price was decreased by 60%. At the end of March, the new price was increased by 60%. At the end of April, the new price was decreased by 60%. On May 1^{st}, the final price was approximately what percent of A?
(A) 41%
(B) 64%
(C) 100%
(D) 136%
(E) 159%
12) Suppose that, at current exchange rates, $1 (US) is equivalent to Q euros, and 1 euro is equivalent to 7Q Chinese Yuan. Suppose that K kilograms of Chinese steel, worth F Chinese Yuan per kilogram, sold to a German company that paid in euros, can be fashioned into N metal frames for chairs. These then are sold to an American company, where plastic seats & backs will be affixed to these frames. If the German company made a total net profit of P euros on this entire transaction, how much did the US company pay in dollars for each frame?
13) At the Zamenhof Language School, at least 70% of the students take English each year, at least 40% take German each year, and between 30% and 60% take Italian each year. Every student must take at least one of these three languages, and no student is allowed to take more than two languages in the same year. What is the possible percentage range for students taking both English and German in the same year?
(A) 0% to 70%
(B) 0% to 100%
(C) 10% to 70%
(D) 10% to 100%
(E) 40% to 70%
14) On any given day, the probability that Bob will have breakfast is more than 0.6. The probability that Bob will have breakfast and will have a sandwich for lunch is less than 0.5. The probability that Bob will have breakfast or will have a sandwich for lunch equals 0.7. Let P = the probability that, on any given day, Bob will have a sandwich for lunch. If all the statements are true, what possible range can be established for P?
(A) 0 < P < 0.6
(B) 0 ≤ P < 0.6
(C) 0 ≤ P ≤ 0.6
(D) 0 < P < 0.7
(E) 0 ≤ P < 0.7
(A) – 64
(B) – 7
(C) 38
(D) 88
(E) 128
Explanations for this problem are at the end of this article.
Here are twenty-eight other articles on this blog with free GMAT Quant practice questions. Some have easy questions, some have medium, and few have quite challenging questions.
1) GMAT Geometry: Is It a Square?
2) GMAT Shortcut: Adding to the Numerator and Denominator
3) GMAT Quant: Difficult Units Digits Questions
4) GMAT Quant: Coordinate Geometry Practice Questions
5) GMAT Data Sufficiency Practice Questions on Probability
6) GMAT Quant: Practice Problems with Percents
7) GMAT Quant: Arithmetic with Inequalities
8) Difficult GMAT Counting Problems
9) Difficult Numerical Reasoning Questions
10) Challenging Coordinate Geometry Practice Questions
11) GMAT Geometry Practice Problems
12) GMAT Practice Questions with Fractions and Decimals
13) Practice Problems on Powers and Roots
14) GMAT Practice Word Problems
15) GMAT Practice Problems: Sets
16) GMAT Practice Problems: Sequences
17) GMAT Practice Problems on Motion
18) Challenging GMAT Problems with Exponents and Roots
19) GMAT Practice Problems on Coordinate Geometry
20) GMAT Practice Problems: Similar Geometry Figures
20) GMAT Practice Problems: Variables in the Answer Choices
21) Counting Practice Problems for the GMAT
22) GMAT Math: Weighted Averages
23) GMAT Data Sufficiency: More Practice Questions
24) Intro to GMAT Word Problems, Part I
25) GMAT Data Sufficiency Geometry Practice Questions
26) GMAT Data Sufficiency Logic: Tautological Questions
27) GMAT Quant: Rates and Ratios
28) Absolute Value Inequalities
These are hard problems. When you read the solutions, don’t merely read them passively. Study the strategies used, and do what you can to retain them. Learn from your mistakes!
1) We need sets of three distinct integers {a, b, c} that have a sum of one-digit number d. There are seven possibilities:
For each set, the sum-digit has to be in the one’s place, but the other three digits can be permutated in 3! = 6 ways in the other three digits. Thus, for each item on that list, there are six different possible four-digit numbers. The total number of possible four-digit numbers would be 7*6 = 42. Answer = (E)
2) The fact that abcd is odd means that cd must be an odd number and that a & b both must be odd. That limits the choices significantly. We know that neither a nor b can equal 1, because any single digit number times 1 is another single digit number, and we need a two-digit product—there are no zeros in abcd. We also know that neither a nor b can equal 5, because any odd multiple of 5 ends in 5, and we would have a repeated digit: the requirement is that all four digits be distinct.
Therefore, for possible values for a & b, we are limited to three odd digits {3, 7, 9}. We can take three different pairs, and in each pair, we can swap the order of a & b. Possibilities:
Those six are the only possibilities for abcd.
Answer = (B)
3) Total number of cars = 500
2D cars total = 165, so
4D cars total = 335
120 4D cars have BUC
“Eighteen percent of all the cars with back-up cameras have standard transmission.”
18% = 18/100 = 9/50
This means that the number of cars with BUC must be a multiple of 50.
How many 2D cars can we add to 120 4D cars to get a multiple of 50? We could add 30, or 80, or 130, but after that, we would run out of 2D cars. These leaves three possibilities for the total number with BUC:
If a total of 150 have BUC, then 18% or 27 of them also have ST.
If a total of 200 have BUC, then 18% or 36 of them also have ST.
If a total of 250 have BUC, then 18% or 45 of them also have ST.
Then we are told: “40% of all the cars with both back-up cameras and standard transmission are two-door car.”
40% = 40/100 = 2/5
This means that number of cars with both back-up cameras and standard transmission must be divisible by 5. Of the three possibilities we have, only the third words.
Total cars with BUC cams = 250 (120 with 4D and 130 with 2D)
18% or 45 of these also have ST.
40% of that is 18, the number of 2D cars with both BUC and ST.
Thus, the number of 4D cars with both BUC and ST would be
45 – 18 = 27
Answer = (B)
4) 700 student total
4G = total number of fourth graders
5G = total number of fifth graders
We are told 4G = 320, so 5G = 700 – 320 = 380
5GM, 5GF = fifth grade boys and girls, respectively
We are told 5GF = 210, so 5GM = 380 – 210 = 170
4GC, 5GC = total number of 4^{th} or 5^{th} graders, respectively taking Chinese
We are told
5GC = 0.5(5G) = 0.5(380) = 190
4GC = 0.4(4G) = 0.4(320) = 128
4GFM, 4GMC, 5GFC, 5GMC = 4^{th}/5^{th} grade boys & girls taking Chinese
We are told that, of the 170 fifth grade boys, 90 do not take Chinese, so 170 = 90 = 80 do. Thus 5GMC = 80.
5GMC + 5GFC = 5GC
80 + 5GFC = 190
5GFC = 110
We are told:
4GFM < (0.5)(5GFC)
4GFM < (0.5)(100)
4GFM < 55
Thus, 4GFM could be as low as zero or as high as 54.
4GMC = 4GC – 4GFM
If 4GFM = 0, then 4GMC = 128 – 0 = 128
If 4GFM = 54, then 4GMC = 128 – 54 = 74
Thus, fourth grade boys taking Mandarin Chinese could take on any value N, such that 74 ≤ N ≤ 128. Of the answer choices listed, the only one that works is 100.
Answer = (D)
5) The single cube has paint on all six sides. Each of the eight boxes in the 2 x 2 x 2 cube has paint on three sides (8 corner pieces). In the 3 x 3 x 3 cube, there are 8 corner pieces, 12 edge pieces (paint on two sides), 6 face pieces (paint on one side), and one interior piece (no paint). In the 4 x 4 x 4 cube, there are 8 corner pieces, 24 edge pieces, 24 face pieces, and 8 interior pieces. This chart summarizes what we have:
For the 10 x 10 flat square, we will need 4 corner pieces that have paint on three sides, 32 edge pieces that have paint on two sides (top & side), and 64 middle pieces that have paint on one side (the top).
We could use either the single total box or any of the 24 corner boxes for the four corners of the square. That leaves 21 of these, and 36 edge boxes, more than enough to cover the 32 edges of the square. The remaining ones, as well as all 30 face boxes, can be turned paint-side-up to fill in the center. The only boxes that will need to be painted, one side each, are the 9 interior boxes. Thus, we have 9 sides to paint.
Answer = (C)
6) Here’s a diagram.
First, let’s count the equilateral triangles. They are {AEI, BFJ, CGK, DHL}. There are only four of them.
Now, consider all possible isosceles triangles, excluding equilateral triangles, with point A as the vertex. We could have BAL, CAK, DAJ, and FAH. All four of those have a line of symmetry that is vertical (through A and G). Thus, we could make those same four triangles with any other point as the vertex, and we would never repeat the same triangle in the same orientation. That’s 4*12 = 48 of these triangles, plus the 4 equilaterals, is 52 total triangles.
Answer = (B)
7) There are five basic scenarios for this:
Case I: (make)(make)(make)(make)(any)
If she makes the first four, then it doesn’t matter if she makes or misses the fifth!
Case II: (miss)(make)(make)(make)(make)
Case III: (make)(miss)(make)(make)(make)
Case IV: (make)(make)(miss)(make)(make)
Case V: (make)(make)(make)(miss)(make)
Put in the probabilities:
Case I: (0.6)(0.8)(0.8)(0.8)
Case II: (0.4)(0.4)(0.8)(0.8)(0.8)
Case III: (0.6)(0.2)(0.4)(0.8)(0.8)
Case IV: (0.6)(0.8)(0.2)(0.4)(0.8)
Case V: (0.6)(0.8)(0.8)(0.2)(0.4)
Since all the answers are fractions, change all of those to fractions. Multiply the first by (5/5) so it has the same denominator as the other products.
Case I: (3/5)(4/5)(4/5)(4/5)(5/5) = 960/5^5
Case II: (2/5)(2/5)(4/5)(4/5)(4/5) = 256/5^5
Case III: (3/5)(1/5)(2/5)(4/5)(4/5) = 96/5^5
Case IV: (3/5)(4/5)(1/5)(2/5)(4/5) = 96/5^5
Case V: (3/5)(4/5)(4/5)(1/5)(2/5) = 96/5^5
Add the numerators. Since 96 = 100 – 4, 3*96 = 3(100 – 4) = 300 – 12 = 288.
288 + 256 + 960 = 1504
P = 1504/5^5
Answer = (E)
8) There are three cases: AABC, ABBC, and ABCC.
In case I, AABC, there are nine choices for A (because A can’t be zero), then 9 for B, then 8 for C. 9*9*8 = 81*8 = 648.
In case II, ABBC, there are 9 choices for A, 9 for B, and 8 for C. Again, 648.
In case III, ABCC, there are 9 choices for A, 9 for B, and 8 for C. Again, 648.
48*3 = (50 – 2)*3 = 150 – 6 = 144
3*648 = 3(600 + 48) = 1800 + 144 = 1948
Answer = (D)
9)
We know that the distance from A (0,6) to B (0, – 6) is 12, so triangle ABO is equilateral. This means that angle AOB is 60°. The entire circle has an area of
A 60° angle is 1/6 of the circle, so the area of sector AOB (the “slice of pizza” shape) is
The area of an equilateral triangle with side s is
Equilateral triangle AOB has s = 12, so the area is
If we subtract the equilateral triangle from the sector, we get everything to the right of the x-axis.
Again, that’s everything to the right of the x-axis, the parts of the circle that lie in Quadrants I & IV. We just want the part in Quadrant I, which would be exactly half of this.
Answer = (C)
10) One point is (50, 70) and one is (100, 89): the line has to pass above both of those. Well, round the second up to (100, 90)—if the line goes above (100, 90), then it definitely goes about (100, 89)!
What is the slope from (50, 70) to (100, 90)? Well, the rise is 90 – 70 = 20, and the run is 100 – 50 = 50, so the slope is rise/run = 20/50 = 2/5. A line with a slope of 2/5 could pass just above these points.
Now, what about the third point? For the sake of argument, let’s say that the line has a slope of 2/5 and goes through the point (50, 71), so it will pass above both of the first two points. Now, move over 5, up 2: it would go through (55, 73), then (60, 75), then (65, 77), then (70, 79), then (75, 81), then (80, 83). This means it would pass under the third point, (80, 84). A slope of 2/5 works for all three points.
We don’t have to do all the calculations, but none of the other slope values works.
Answer = (D)
11) The trap answer is 100%: a percent increase and percent decrease by the same percent do not cancel out.
Let’s say that the A = $100 at the beginning of the year.
End of January, 60% increase. New price = $160
End of February, 60% decrease: that’s a decrease of 60% of $160, so that only 40% of $160 is left.
10% of $160 = $16
40% of $160 = 4(16) = $64
That’s the price at the end of February.
End of March, a 60% increase: that’s a increase of 60% of $64.
10% of $64 = $6.40
60% of $64 = 6(6 + .40) = 36 + 2.4 = $38.40
Add that to the starting amount, $64:
New price = $64 + $38.40 = $102.40
End of April, 60% decrease: that’s a decrease of 60% of $102.40, so that only 40% of $102.40 is left.
At this point, we are going to approximate a bit. Approximate $102.40 as $100, so 40% of that would be $40. The final price will be slightly more than $40.
Well, what is slightly more than $40, as a percent of the beginning of the year price of $100? That would be slightly more than 40%.
Answer = (A)
12) The K kilograms, worth F Chinese Yuan per kilogram, are worth a total of KF Chinese Yuan. The German company must pay this amount.
Since 1 euro = (7Q) Chinese Yuan, then (1/(7Q)) euro = 1 Chinese Yuan, and (KF/7Q) euros = KF Chinese Yuan. That’s the amount that the Germans pay to the Chinese.
That is the German company’s outlay, in euros. Now, they make N metal chairs, and sell them, making a gross profit of P euros.
That must be the total revenue of the German company, in euros. This comes from the sale to the American company. Since $1 = Q euros, $(1/Q) = 1 euro, so we change that entire revenue expression to euros to dollars, we divide all terms by Q.
That must be the total dollar amount that leaves the American company and goes to the German company. This comes from the sale of N metal frames for chairs, so each one must have been 1/N of that amount.
Answer = (A)
13) First, we will focus on the least, the lowest value. Suppose the minimum of 70% take English, and the minimum of 40% take German. Even if all 30% of the people not taking English take German, that still leaves another 10% of people taking German who also have to be taking English. Thus, 10% is the minimum of this region.
Now, the maximum. Both the German and English percents are “at least” percents, so either could be cranked up to 100%. The trouble is, though, that both can’t be 100%, because some folks have to take Italian, and nobody can take three languages at once. The minimum taking Italian is 30%. Let’s assume all 100% take German, and that everyone not taking Italian is taking English: that’s 70% taking English, all of whom also would be taking German. Thus, 70% is the maximum of this region.
Answer = (C)
14) Let A = Bob eats breakfast, and B = Bob has a sandwich for lunch. The problem tells us that:
P(A) > 0.6
P(A and B) < 0.5
P(A or B) = 0.7
First, let’s establish the minimum value. If Bob never has a sandwich for lunch, P(B) = 0, then it could be that P(A and B) = 0, which is less than 0.5, and it could be that P(A) = 0.7, which is more than 0.6, so that P(A or B) = 0.7. All the requirements can be satisfied if P(B) = 0, so it’s possible to equal that minimum value.
Now, the maximum value. Since P(A or B) = 0.7, both P(A) and P(B) must be contained in this region. See the conceptual diagram.
The top line, 1, is the entire probability space. The second line, P(A or B) = 0.7, fixes the boundaries for A and B. P(A) is the purple arrow, extending from the right. P(B) is the green arrow extending from the left. The bottom line, P(A and B) < 0.5, is the constraint on their possible overlap.
Let’s say that P(A) is just slightly more than 0.6. That means the region outside of P(A), but inside of P(A or B) is slightly less than 1. That’s the part of P(B) that doesn’t overlap with P(A). Then, the overlap has to be less than 0.5. If we add something less than 1 to something less than 5, we get something less than 6. P(B) can’t equal 0.6, but it can any value arbitrarily close to 0.6.
Thus, 0 ≤ P(B) < 0.6.
Answer = (B)
15)
Answer = (E)
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]]>1) Seven children — A, B, C, D, E, F, and G — are going to sit in seven chairs in a row. The children C & F have to sit next to each other, and the others can sit in any order in any remaining chairs. How many possible configurations are there for the children?
2) Seven children — A, B, C, D, E, F, and G — are going to sit in seven chairs in a row. Child A has to sit next to both B & G, with these two children immediately adjacent to here on either side. The other four children can sit in any order in any of the remaining seats. How many possible configurations are there for the children?
3) Six children — A, B, C, D, E, and F — are going to sit in six chairs in a row. Child E must be somewhere to the left of child F. How many possible configurations are there for the children?
4) Seven children — A, B, C, D, E, F, and G — are going to sit in seven chairs in a row. Children A & B must sit next to each other, and child C must be somewhere to the right of A & B. How many possible configurations are there for the children?
If these problems make your head spin, then you have found the right post!
First of all, it’s vitally important to understand something called the Fundamental Counting Principle (FCP). It’s also important to understand permutations and combinations. These ideas underlie all the introductory counting problems on the GMAT Quantitative section, and they form the basis of what we will discuss in these posts.
Whenever you are counting arrangements or possibilities in a complex scenario, always start with the most restrictive feature or element. Quite often, the plan is — (a) count all the possibilities for the elements with restrictions; (b) count all the possibilities for the remaining non-restricted items; (c) by the FCP, multiply those numbers together.
For example, let’s take a simple case, a bit simpler than anything the GMAT will throw your way. Suppose we have five children, A, B, C, D, and E, and we are going to sit them in five chairs in a row, with the restriction that B & C have to be next to each other.
Step #1: consider the restricted elements, B & C. They have to be next to each other. How many “next to each other pairs are there among the four seats?
X X _ _ _
_ X X _ _
_ _ X X _
_ _ _ X X
There are those four possible pairs. Now, B & C can be in either order in any of those, so that’s 4*2 = 8 possibilities for these two children.
Step #2: now, the non-restricted elements. Once B & C are seated, we can put A & D & E in any order in the remaining three seats. Three elements in any order — that’s a permutation — 3! = 6. For any particular configuration of B & C, the remaining three can be seated in 6 different ways.
Step #3: combine with the FCP. Children B & C can be seated in 8 different configurations, and for each one of those, A & D & E can be seated in 6 different configurations. Total number of configurations = 6*8 = 48.
That’s the basic process, which can be expanded for larger numbers of items. Having seen that, you may want to give the problems above a second look before reading through the explanations below. If those problems still are confounding for you, read through the solutions below very carefully. If you thoroughly understand the four problems at the top of this post, you will be able to handle anything counting problem GMAT will give you.
1) First, we will consider the restricted elements — C & F. How many ways can they sit next to each other in a row of seven chairs? Well, first of all, how many “next to each other” pairs of chairs are there?
X X _ _ _ _ _
_ X X _ _ _ _
_ _ X X _ _ _
_ _ _ X X _ _
_ _ _ _ X X _
_ _ _ _ _ X X
There are six different pairs of “next to each other” chairs. For each pair, children C & F could be in either order, so that’s 6*2 = 12 possibilities for these two.
Now, consider the other five children. For any configuration of C & F, the remaining five children could be seated in any order among the five remaining seats. Five items in any order — that’s a combination of the 5 items —- 5P5 = 5! = 120. For any single configuration of C & F, there are 120 ways that the other children could be seated in the remaining seats.
Finally, we’ll combine with the Fundamental Counting Principle. We have 12 ways for the first two, and 120 ways for the remaining five. That’s a total number of configurations of 12*120 = 1440. Answer = C
2) First, we will consider the restricted elements — children A & B & G have to be in three seats in a row. How many “three in a row” seats are there in a row of seven seats?
X X X _ _ _ _
_ X X X _ _ _
_ _ X X X _ _
_ _ _ X X X _
_ _ _ _ X X X
There are five different “three in a row” locations for these three children. Now, for any given triplet of seats, we know A has to be in the middle, so the children could be seated B-A-G or G-A-B — just those two orders. This means the total number of configurations for these three children is 5*2 = 10.
Now, consider the non-restricted elements, the other four. Once A & B & G are seated, the remaining four children can be seated in any order among the remaining four seats — that’s a permutation of the 4 items —- 4P4 = 4! = 24. For any single configuration of A & B & G, there are 24 ways that the other children could be seated in the remaining seats.
Finally, we’ll combine with the Fundamental Counting Principle. We have 10 ways for the first three, and 24 ways for the remaining four. That’s a total number of configurations of 24*10 = 240. Answer = A
3) If we wanted, we could make this one extremely difficult, counting out all kinds of possibilities in several different cases. Instead, we are going to make this ridiculously easy.
First of all, with absolutely no restrictions, how many ways can the six children be arranged on the six chairs? That’s a permutation of the 6 items —- 6P6 = 6! = 720. That’s the total number of arrangements with no restrictions. Of course, those 720 arrangements have all kinds of symmetry to them. In particular, in all of those arrangements overall, it’s just as likely for E to be to the left of F as it is for E to be to the right of F. Therefore, exactly half must have E to the right of F, and exactly half must have E to the left of F. Therefore, exactly (1/2)*720 = 360 of the arrangements have E to the left of F. Answer = D.
4) This problem is a notch harder than anything you are likely to see on the GMAT. If you can master the principles in this problem, you certainly will be able to handle almost any problem the GMAT could concoct.
First, consider the restriction of A & B. As with problem #1 above, there are 12 possibilities for A & B, counting both position and order.
Now, put the other five in any order — that’s 120 possibilities, for a total number of configurations of 1440. That number does not take into account the restriction with C.
Think about those 1440 configurations. In exactly half of them, C will be to the right of A & B, and exactly half, C will be to the left of A & B. Therefore, in (1/2)*1440 = 720 configurations, C will be to the right of A & B. Answer = B
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]]>1) A radio station has to choose three days of the seven in a week to broadcast a certain program, and that set will repeat each week. The program can be broadcast equally on any of the seven weekdays —- weekdays vs. weekends don’t matter at all —- nor does it matter whether the days the program airs are adjacent or not. Absolutely any three of the seven weekdays can be chosen. How many different three-day combinations of the seven weekdays can be constructed?
2) Claudia can choose any two of four different candles and any 8 of 9 different flowers for a centerpiece arrangement. Given these choices, how many candle + flower groupings can she select?
3) A newly-wed couple is using a website to design an eBook Wedding Album to distribute to their friends and families. The template they have chosen has places for 3 large photos and 19 smaller photos. The couple has 6 large photos they could use for those three slots, and 21 smaller photos they could use for those 19 slots. Given these choices, how many different possible albums could they create?
In this post, we’ll discuss how to handle questions like this — without a calculator.
Mathematically, a combination is a group of things, irrespective of order. For example, {A, B, D} and {D, A, B} and {B, A, D} are all the same combination — order doesn’t matter at all. The expression nCr (read “n choose r”) is the expression for the number of combinations of r things, r choices, you can make from a pool of n unique items. For example,
6C3 = the number of combinations of three one can choose from a pool of six unique items.
In a previous post about combinations, I give the following formula for nCr
where the exclamation point (“!”) is the factorial symbol — n! means the product of all the positive integers from n down to 1. Using this formula, we could compute the value of 6C3
So, it turns out, there are twenty ways to pick a set of three items from a pool of six unique items. That’s one way to calculate nCr, but it’s not the only way.
The mathematician and philosopher Blaise Pascal (1623 – 1662) created a magical triangular array of numbers known now as Pascal’s Triangle:
How does this pattern work? Well, of course, the edges are diagonals of 1’s. Every inside number is the sum of the two numbers above it in the previous row, diagonally to the left and diagonally to the right. For example, the 2 is the sum 1+1; both 3’s are the sum 1+2; both 4’s are the sums 1+3; the 6 is the sum 3+3, etc. They often show Pascal’s Triangle to grade school students to give them practice with addition.
Despite its relatively easy origins, Pascal’s Triangle is a treasure trove of miraculous mathematical properties. Most relevant for us right now is: Pascal’s Triangle is, among other things, an array of all possible nCr’s.
nCr = the rth entry of the nth row of Pascal’s Triangle
In that definition, we have to be careful — we have to start counting at zero instead of one. The top 1 on Pascal’s Triangle is the zeroth row, zeroth entry, 0C0 = 1 (a relatively meaningless number in terms of combinations!) The next row (1, 1) is the first row, and the next row is the second row (1, 2, 1), etc. Notice that the second number in any row (as well as the penultimate number in any row) equals the row number. The first number (always 1) is actually the zeroth entry, so that second number would actually be the first entry — the first entry of the nth row always equals n. In other words
nC1 = n
That makes sense: if we have n different items, we have exactly n ways of selecting any one item. Those entries, the first entries of each row, line along a diagonal down the left side of the triangle. Because of the complete symmetry of the triangle, this always equals the numbers on the corresponding diagonals on the right side, which would be the (n-1) entries of each row. Thus:
nC1 = nC(n-1) = n
When you have to figure out nCr when n & r are both relatively small numbers, it may be easier simply to jot down the first few rows of Pascal’s Triangle. For example, with what we have showing, we can see that 6C3, the 3rd entry of the 6th row, is 20 — the same as the answer we found via the factorials formula.
Things get even more interesting when we move to the next diagonal in, shown in green here:
These numbers, the set of the second entries in each row, are the triangular numbers. Among other things, the second entry in the n row is the sum of the first n-1 positive integers. For example
3 = 2 + 1
6 = 3 + 2 + 1
10 = 4 + 3 + 2 + 1 etc.
The formula for this is:
Because we have a formula, we can calculate this for much higher numbers. For 21C2, we would have to write out everything to the twenty-first row of Pascal’s Triangle, an arduous undertaking. Rather, we could simply use the formula
Notice that the symmetry of Pascal’s Triangle also provides tremendous insight into the nature of the nCr numbers. First of all, in any row, the second entry, the triangular number in that row, must be equal to the third-to-last entry of the row, that is, the (n-2) entry of the row. Thus
Thus, via the triangular numbers, we have a formula, not only for the second entry of each row, but also for the third-to-last entry of every row. Thus, it’s very easy to figure out the first three or last three numbers in any row. More generally, symmetry guarantees that:
nCr = nC(n-r)
If you think about combinations this makes sense: if we have a pool of n unique items, then every time we choose a unique set of r items, we necessarily exclude a corresponding unique set of (n-r) items. In other words, there is necessarily a 1-to-1 correspondence between unique sets of r elements and unique sets of the other (n-r) elements —- because there’s a 1-to-1 correspondence, the number of each must always be the same. This is precisely what that equation says.
That discussion was liberally peppered with hints about how to do the above three questions. If you had trouble with them on the first pass, you may want to give them a second look before proceeding to the explanations below. Here’s an additional question from inside Magoosh:
4) http://gmat.magoosh.com/questions/847
1) Behind the story, we are really being asked to evaluate 7C3. We could use the factorial formula, but above we conveniently happen to have Pascal’s Triangle written out to the seventh row. We see that 7C3, the third entry of the seventh row, is 35. Answer = D.
2) For this one, we have to use the Fundamental Counting Principle (FCP) as well as information about combinations. For the flowers, we want 9C8, which by the symmetry of Pascal’s Triangle, has to equal 9C1, the first entry in the row, which of course equals the row number.
9C8 = 9C1 = 9
That’s the number of flower combinations. For the candles, 4C2, we read the second entry of the fourth row of Pascal’s Triangle.
4C2 = 6
Now, by the FCP, we multiply these for the total number of centerpiece arrangements: 6*9 = 54. Answer = A
3) For the large photos, we need 6C3, which we calculated in the article:
6C3 = 20
For the smaller photos, we need 21C19, which by symmetry must equal 21C2, and we have a formula for that. In fact, in the article above, we already calculated that 21C2 = 210.
Now, by the FCP, we just multiply these: total number of possible albums = 20*210 = 4200. Answer = B
The post GMAT Math: Calculating Combinations appeared first on Magoosh GMAT Blog.
]]>The post Breakdown of GMAT Quant Concepts by Frequency appeared first on Magoosh GMAT Blog.
]]>Here are the samples of Official Material he used to figure out the GMAT Quant breakdown:
1.GMAT Official Guide (12th Edition) Problem Solving Practice Questions (Pg. 152-265)*
2. GMAT Official Guide (12th Edition) Data Sufficiency Practice Questions (Pg. 272-351)*
3. GMATPrep Test #1
4. GMATPrep Test #2
5. Released past exam from GMAC, test code 14, from the 90s/early 2000’s
Total | Percentage | ||
Arithmetic | Percents | 21 | 10.71% |
Arithmetic | Properties of Integers | 18 | 9.18% |
Arithmetic | Descriptive Statistics | 12 | 6.12% |
Arithmetic | Fractions | 11 | 5.61% |
Algebra | Linear Equations, Two Unknowns | 11 | 5.61% |
Algebra | Simplifying Algebraic Expressions | 9 | 4.59% |
Arithmetic | Powers & Roots of Numbers | 8 | 4.08% |
Arithmetic | Counting Methods | 7 | 3.57% |
Algebra | Linear Equations, One Unknown | 7 | 3.57% |
Algebra | Functions/Series | 7 | 3.57% |
Geometry | Coordinate Geometry | 7 | 3.57% |
Word Problems | Rate Problems | 7 | 3.57% |
Arithmetic | Ratio & Proportions | 6 | 3.06% |
Arithmetic | Decimals | 5 | 2.55% |
Arithmetic | Discrete Probability | 5 | 2.55% |
Algebra | Exponents | 5 | 2.55% |
Algebra | Inequalities | 5 | 2.55% |
Geometry | Quadrilaterals | 5 | 2.55% |
Geometry | Circles | 5 | 2.55% |
Word Problems | Interest Problems | 4 | 2.04% |
Algebra | Solving Quadratic Equations | 3 | 1.53% |
Geometry | Triangles | 3 | 1.53% |
Geometry | Rectangular Solids & Cylinders | 3 | 1.53% |
Word Problems | Work Problems | 3 | 1.53% |
Word Problems | Mixture Problems | 3 | 1.53% |
Geometry | Intersecting Angles and Lines | 2 | 1.02% |
Word Problems | Profit | 2 | 1.02% |
Word Problems | Sets | 2 | 1.02% |
Word Problems | Measurement Problems | 2 | 1.02% |
Word Problems | Data Interpretation | 2 | 1.02% |
Arithmetic | Real Numbers | 1 | 0.51% |
Arithmetic | Sets | 1 | 0.51% |
Algebra | Solving by Factoring | 1 | 0.51% |
Geometry | Polygons | 1 | 0.51% |
Word Problems | Discount | 1 | 0.51% |
Word Problems | Geometry Problems | 1 | 0.51% |
Algebra | Equations | 0 | 0.00% |
Algebra | Absolute Value | 0 | 0.00% |
Geometry | Lines | 0 | 0.00% |
Geometry | Perpendicular Lines | 0 | 0.00% |
Geometry | Parallel Lines | 0 | 0.00% |
Total | Percentage | ||
Arithmetic | Properties of Integers | 30 | 19.23% |
Arithmetic | Descriptive Statistics | 13 | 8.33% |
Arithmetic | Percents | 11 | 7.05% |
Algebra | Linear Equations, Two Unknowns | 11 | 7.05% |
Word Problems | Rate Problems | 9 | 5.77% |
Algebra | Inequalities | 8 | 5.13% |
Arithmetic | Sets | 7 | 4.49% |
Arithmetic | Ratio & Proportions | 6 | 3.85% |
Geometry | Triangles | 6 | 3.85% |
Geometry | Circles | 6 | 3.85% |
Arithmetic | Decimals | 5 | 3.21% |
Geometry | Coordinate Geometry | 5 | 3.21% |
Arithmetic | Fractions | 4 | 2.56% |
Algebra | Linear Equations, One Unknown | 4 | 2.56% |
Algebra | Exponents | 4 | 2.56% |
Geometry | Rectangular Solids & Cylinders | 4 | 2.56% |
Arithmetic | Discrete Probability | 3 | 1.92% |
Algebra | Functions/Series | 3 | 1.92% |
Word Problems | Interest Problems | 3 | 1.92% |
Geometry | Lines | 2 | 1.28% |
Word Problems | Work Problems | 2 | 1.28% |
Word Problems | Discount | 2 | 1.28% |
Word Problems | Profit | 2 | 1.28% |
Arithmetic | Real Numbers | 1 | 0.64% |
Arithmetic | Counting Methods | 1 | 0.64% |
Algebra | Simplifying Algebraic Expressions | 1 | 0.64% |
Algebra | Absolute Value | 1 | 0.64% |
Geometry | Quadrilaterals | 1 | 0.64% |
Word Problems | Measurement Problems | 1 | 0.64% |
Arithmetic | Powers & Roots of Numbers | 0 | 0.00% |
Algebra | Equations | 0 | 0.00% |
Algebra | Solving by Factoring | 0 | 0.00% |
Algebra | Solving Quadratic Equations | 0 | 0.00% |
Geometry | Intersecting Angles and Lines | 0 | 0.00% |
Geometry | Perpendicular Lines | 0 | 0.00% |
Geometry | Parallel Lines | 0 | 0.00% |
Geometry | Polygons | 0 | 0.00% |
Word Problems | Mixture Problems | 0 | 0.00% |
Word Problems | Sets | 0 | 0.00% |
Word Problems | Geometry Problems | 0 | 0.00% |
Word Problems | Data Interpretation | 0 | 0.00% |
Total | Percentage | ||
Arithmetic | Properties of Integers | 6 | 17.14% |
Arithmetic | Ratio & Proportions | 4 | 11.43% |
Arithmetic | Descriptive Statistics | 4 | 11.43% |
Algebra | Exponents | 3 | 8.57% |
Algebra | Functions/Series | 3 | 8.57% |
Arithmetic | Percents | 2 | 5.71% |
Arithmetic | Powers & Roots of Numbers | 2 | 5.71% |
Word Problems | Interest Problems | 2 | 5.71% |
Arithmetic | Fractions | 1 | 2.86% |
Arithmetic | Counting Methods | 1 | 2.86% |
Algebra | Equations | 1 | 2.86% |
Geometry | Intersecting Angles and Lines | 1 | 2.86% |
Geometry | Triangles | 1 | 2.86% |
Geometry | Circles | 1 | 2.86% |
Word Problems | Work Problems | 1 | 2.86% |
Word Problems | Sets | 1 | 2.86% |
Word Problems | Measurement Problems | 1 | 2.86% |
Arithmetic | Decimals | 0 | 0.00% |
Arithmetic | Real Numbers | 0 | 0.00% |
Arithmetic | Sets | 0 | 0.00% |
Arithmetic | Discrete Probability | 0 | 0.00% |
Algebra | Simplifying Algebraic Expressions | 0 | 0.00% |
Algebra | Linear Equations, One Unknown | 0 | 0.00% |
Algebra | Linear Equations, Two Unknowns | 0 | 0.00% |
Algebra | Solving by Factoring | 0 | 0.00% |
Algebra | Solving Quadratic Equations | 0 | 0.00% |
Algebra | Inequalities | 0 | 0.00% |
Algebra | Absolute Value | 0 | 0.00% |
Geometry | Lines | 0 | 0.00% |
Geometry | Perpendicular Lines | 0 | 0.00% |
Geometry | Parallel Lines | 0 | 0.00% |
Geometry | Polygons | 0 | 0.00% |
Geometry | Quadrilaterals | 0 | 0.00% |
Geometry | Rectangular Solids & Cylinders | 0 | 0.00% |
Geometry | Coordinate Geometry | 0 | 0.00% |
Word Problems | Rate Problems | 0 | 0.00% |
Word Problems | Mixture Problems | 0 | 0.00% |
Word Problems | Discount | 0 | 0.00% |
Word Problems | Profit | 0 | 0.00% |
Word Problems | Geometry Problems | 0 | 0.00% |
Word Problems | Data Interpretation | 0 | 0.00% |
Total | Percentage | ||
Arithmetic | Descriptive Statistics | 6 | 18.75% |
Arithmetic | Properties of Integers | 4 | 12.50% |
Arithmetic | Ratio & Proportions | 2 | 6.25% |
Arithmetic | Percents | 2 | 6.25% |
Algebra | Simplifying Algebraic Expressions | 2 | 6.25% |
Algebra | Exponents | 2 | 6.25% |
Geometry | Coordinate Geometry | 2 | 6.25% |
Arithmetic | Fractions | 1 | 3.13% |
Arithmetic | Powers & Roots of Numbers | 1 | 3.13% |
Arithmetic | Counting Methods | 1 | 3.13% |
Algebra | Equations | 1 | 3.13% |
Algebra | Linear Equations, Two Unknowns | 1 | 3.13% |
Algebra | Inequalities | 1 | 3.13% |
Algebra | Functions/Series | 1 | 3.13% |
Geometry | Triangles | 1 | 3.13% |
Geometry | Circles | 1 | 3.13% |
Word Problems | Rate Problems | 1 | 3.13% |
Word Problems | Discount | 1 | 3.13% |
Word Problems | Measurement Problems | 1 | 3.13% |
Arithmetic | Decimals | 0 | 0.00% |
Arithmetic | Real Numbers | 0 | 0.00% |
Arithmetic | Sets | 0 | 0.00% |
Arithmetic | Discrete Probability | 0 | 0.00% |
Algebra | Linear Equations, One Unknown | 0 | 0.00% |
Algebra | Solving by Factoring | 0 | 0.00% |
Algebra | Solving Quadratic Equations | 0 | 0.00% |
Algebra | Absolute Value | 0 | 0.00% |
Geometry | Lines | 0 | 0.00% |
Geometry | Intersecting Angles and Lines | 0 | 0.00% |
Geometry | Perpendicular Lines | 0 | 0.00% |
Geometry | Parallel Lines | 0 | 0.00% |
Geometry | Polygons | 0 | 0.00% |
Geometry | Quadrilaterals | 0 | 0.00% |
Geometry | Rectangular Solids & Cylinders | 0 | 0.00% |
Word Problems | Work Problems | 0 | 0.00% |
Word Problems | Mixture Problems | 0 | 0.00% |
Word Problems | Interest Problems | 0 | 0.00% |
Word Problems | Profit | 0 | 0.00% |
Word Problems | Sets | 0 | 0.00% |
Word Problems | Geometry Problems | 0 | 0.00% |
Word Problems | Data Interpretation | 0 | 0.00% |
Total | Percentage | ||
Arithmetic | Properties of Integers | 7 | 13.73% |
Arithmetic | Ratio & Proportions | 5 | 9.80% |
Arithmetic | Percents | 5 | 9.80% |
Arithmetic | Descriptive Statistics | 4 | 7.84% |
Word Problems | Sets | 4 | 7.84% |
Arithmetic | Powers & Roots of Numbers | 3 | 5.88% |
Algebra | Linear Equations, Two Unknowns | 3 | 5.88% |
Word Problems | Rate Problems | 3 | 5.88% |
Arithmetic | Fractions | 2 | 3.92% |
Arithmetic | Counting Methods | 2 | 3.92% |
Algebra | Exponents | 2 | 3.92% |
Algebra | Inequalities | 2 | 3.92% |
Geometry | Triangles | 2 | 3.92% |
Algebra | Simplifying Algebraic Expressions | 1 | 1.96% |
Algebra | Equations | 1 | 1.96% |
Algebra | Functions/Series | 1 | 1.96% |
Geometry | Quadrilaterals | 1 | 1.96% |
Geometry | Circles | 1 | 1.96% |
Geometry | Rectangular Solids & Cylinders | 1 | 1.96% |
Geometry | Coordinate Geometry | 1 | 1.96% |
Arithmetic | Decimals | 0 | 0.00% |
Arithmetic | Real Numbers | 0 | 0.00% |
Arithmetic | Sets | 0 | 0.00% |
Arithmetic | Discrete Probability | 0 | 0.00% |
Algebra | Linear Equations, One Unknown | 0 | 0.00% |
Algebra | Solving by Factoring | 0 | 0.00% |
Algebra | Solving Quadratic Equations | 0 | 0.00% |
Algebra | Absolute Value | 0 | 0.00% |
Geometry | Lines | 0 | 0.00% |
Geometry | Intersecting Angles and Lines | 0 | 0.00% |
Geometry | Perpendicular Lines | 0 | 0.00% |
Geometry | Parallel Lines | 0 | 0.00% |
Geometry | Polygons | 0 | 0.00% |
Word Problems | Work Problems | 0 | 0.00% |
Word Problems | Mixture Problems | 0 | 0.00% |
Word Problems | Interest Problems | 0 | 0.00% |
Word Problems | Discount | 0 | 0.00% |
Word Problems | Profit | 0 | 0.00% |
Word Problems | Geometry Problems | 0 | 0.00% |
Word Problems | Measurement Problems | 0 | 0.00% |
Word Problems | Data Interpretation | 0 | 0.00% |
Total | Percentage | ||
Arithmetic | Properties of Integers | 65 | 13.83% |
Fractions | 19 | 4.04% | |
Decimals | 10 | 2.13% | |
Real Numbers | 2 | 0.43% | |
Ratio & Proportions | 23 | 4.89% | |
Percents | 41 | 8.72% | |
Powers & Roots of Numbers | 14 | 2.98% | |
Descriptive Statistics | 39 | 8.30% | |
Sets | 8 | 1.70% | |
Counting Methods | 12 | 2.55% | |
Discrete Probability | 8 | 1.70% | |
Algebra | Simplifying Algebraic Expressions | 13 | 2.77% |
Equations | 3 | 0.64% | |
Linear Equations, One Unknown | 11 | 2.34% | |
Linear Equations, Two Unknowns | 26 | 5.53% | |
Solving by Factoring | 1 | 0.21% | |
Solving Quadratic Equations | 3 | 0.64% | |
Exponents | 16 | 3.40% | |
Inequalities | 16 | 3.40% | |
Absolute Value | 1 | 0.21% | |
Functions/Series | 15 | 3.19% | |
Geometry | Lines | 2 | 0.43% |
Intersecting Angles and Lines | 3 | 0.64% | |
Perpendicular Lines | 0 | 0.00% | |
Parallel Lines | 0 | 0.00% | |
Polygons | 1 | 0.21% | |
Triangles | 13 | 2.77% | |
Quadrilaterals | 7 | 1.49% | |
Circles | 14 | 2.98% | |
Rectangular Solids & Cylinders | 8 | 1.70% | |
Coordinate Geometry | 15 | 3.19% | |
Word Problems | Rate Problems | 20 | 4.26% |
Work Problems | 6 | 1.28% | |
Mixture Problems | 3 | 0.64% | |
Interest Problems | 9 | 1.91% | |
Discount | 4 | 0.85% | |
Profit | 4 | 0.85% | |
Sets | 7 | 1.49% | |
Geometry Problems | 1 | 0.21% | |
Measurement Problems | 5 | 1.06% | |
Data Interpretation | 2 | 0.43% | |
Total | 470 | 100.00% |
Total | Percentage | ||
Arithmetic | Properties of Integers | 65 | 13.83% |
Arithmetic | Percents | 41 | 8.72% |
Arithmetic | Descriptive Statistics | 39 | 8.30% |
Algebra | Linear Equations, Two Unknowns | 26 | 5.53% |
Arithmetic | Ratio & Proportions | 23 | 4.89% |
Word Problems | Rate Problems | 20 | 4.26% |
Arithmetic | Fractions | 19 | 4.04% |
Algebra | Exponents | 16 | 3.40% |
Algebra | Inequalities | 16 | 3.40% |
Algebra | Functions/Series | 15 | 3.19% |
Geometry | Coordinate Geometry | 15 | 3.19% |
Arithmetic | Powers & Roots of Numbers | 14 | 2.98% |
Geometry | Circles | 14 | 2.98% |
Algebra | Simplifying Algebraic Expressions | 13 | 2.77% |
Geometry | Triangles | 13 | 2.77% |
Arithmetic | Counting Methods | 12 | 2.55% |
Algebra | Linear Equations, One Unknown | 11 | 2.34% |
Arithmetic | Decimals | 10 | 2.13% |
Word Problems | Interest Problems | 9 | 1.91% |
Arithmetic | Sets | 8 | 1.70% |
Arithmetic | Discrete Probability | 8 | 1.70% |
Geometry | Rectangular Solids & Cylinders | 8 | 1.70% |
Geometry | Quadrilaterals | 7 | 1.49% |
Word Problems | Sets | 7 | 1.49% |
Word Problems | Work Problems | 6 | 1.28% |
Word Problems | Measurement Problems | 5 | 1.06% |
Word Problems | Discount | 4 | 0.85% |
Word Problems | Profit | 4 | 0.85% |
Algebra | Equations | 3 | 0.64% |
Algebra | Solving Quadratic Equations | 3 | 0.64% |
Geometry | Intersecting Angles and Lines | 3 | 0.64% |
Word Problems | Mixture Problems | 3 | 0.64% |
Arithmetic | Real Numbers | 2 | 0.43% |
Geometry | Lines | 2 | 0.43% |
Word Problems | Data Interpretation | 2 | 0.43% |
Algebra | Solving by Factoring | 1 | 0.21% |
Algebra | Absolute Value | 1 | 0.21% |
Geometry | Polygons | 1 | 0.21% |
Word Problems | Geometry Problems | 1 | 0.21% |
Geometry | Perpendicular Lines | 0 | 0.00% |
Geometry | Parallel Lines | 0 | 0.00% |
The list of concepts tested on the quantitative section is from GMAC, which you can see on page 107 of the Official Guide (either 12th or 13th edition). It isn’t perfect– “Integer Properties” is a wide area of knowledge, whereas something like “Circles” is very specific.
NO. For the sake of simplicity and accuracy in reporting absolute frequency, we’ve only assigned each question to one concept. This means that even though GMAC lists “Perpendicular lines” as a topic tested on the GMAT, and we have 0 questions marked as pertaining to that topic, that certainly doesn’t mean the idea of perpendicular lines did not come up at all on all of the exams. It certainly appeared, but often in questions that were better categorized, overall, as “Coordinate Geometry”, or “Intersecting Angles and Lines”.
We hope this serves as a guideline for the relative frequency of math topics tested on the GMAT to help you decide how to focus your time! In Magoosh practice, you can set up customized practice sessions to focus on specific concepts, as well as review your performance on individual concepts to identify your weak spots using our Review tool.
Let us know whether you find this type of breakdown helpful, and whether you have any questions about any of the information above! 🙂
The post Breakdown of GMAT Quant Concepts by Frequency appeared first on Magoosh GMAT Blog.
]]>The post GMAT Permutations and Combinations appeared first on Magoosh GMAT Blog.
]]>A permutation is a possible order in which to put a set of objects. Suppose I had a shelf of 5 different books, and I wanted to know: in how many different orders can I put these 5 books? Another way to say that is: 5 books have how many different permutations?
In order to answer this question, we need an odd math symbol: the factorial. It’s written as an exclamation sign, and it means: the product of that number and all the positive integers below it, down to 1. For example, 4! (read “four factorial“) is
4! = (4)(3)(2)(1) = 24
Here’s the permutation formula:
# of permutations of n objects = n!
So, five books the number of permutations is 5! = (5)(4)(3)(2)(1) = 120
A combination is a selection from a larger set. Suppose there is a class of 20, and we are going to pick a team of three people at random, and we want to know: how many different possible three-person teams could we pick? Another way to say that is: how many different combinations of 3 can be taken from a set of 20?
This formula is scary looking, but really not bad at all. If n is the size of the larger collection, and r is the number of elements that will be selected, then the number of combinations is given by
# of combinations =
Again, this looks complicated, but it gets simple very fast. In the question just posed, n = 20, r = 3, and n – r = 17. Therefore,
# of combinations =
To simplify this, consider that:
20! = (20)(19)(18)(17)(the product of all the numbers less than 17)
Or, in other words,
20! = (20)(19)(18)(17!)
That neat little trick allow us to enormously simplify the combinations formula:
# of combinations =
That example is most likely harder than anything you’ll see on the GMAT math, but you may be asked to find combinations with smaller numbers.
1) A bookseller has two display windows. She plans to display 4 new fiction books in the left window, and 3 new non-fiction books in the right window. Assuming she can put the four fiction books in any order, and separately, the three non-fiction books in any order, how many total configurations will there be for the two display windows?
2) The county-mandated guidelines at a certain community college specify that for the introductory English class, the professor may choose one of three specified novels, and choose two from a list of 5 specified plays. Thus, the reading list for this introductory class is guaranteed to have one novel and two plays. How many different reading lists could a professor create within these parameters?
1) The left window will have permutations of the 4 fiction books, so the number of possibilities for that window is
permutations = 4! = (4)(3)(2)(1) = 24
The right window will have permutations of the 3 non-fiction books, so the number of possibilities for that window is
permutations = 3! = (3)(2)(1) = 6
Any of the 24 displays of the left window could be combined with any of the 6 displays of the right window, so the total number of configurations is 24*6 = 144
Answer: C.
2) There are three possibilities for the novel. With the plays, we are taken a combination of 2 from a set of 5 n = 5, r = 2, n – r = 3
# of combinations = =
If the plays are P, Q, R, S, and T, then the 10 sets of two are PQ, PR, PS, PT, QR, QS, QT, RS, RT, & ST.
Any of the three novels can be grouped with any of the 10 possible pairs of plays, for a total of 30 possible reading lists.
Answer: B.
To find out where permutations and combinations sit in the “big picture” of GMAT Quant, and what other Quant concepts you should study, check out our post entitled:
What Kind of Math is on the GMAT? Breakdown of Quant Concepts by Frequency
The post GMAT Permutations and Combinations appeared first on Magoosh GMAT Blog.
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