First, try these challenging GMAT Quantitative problems, all variations on a theme, as you will see.
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?

(A) 600
(B) 720
(C) 1440
(D) 4320
(E) 4800
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?

(A) 240
(B) 480
(C) 720
(D) 1440
(E) 3600
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?

(A) 60
(B) 180
(C) 240
(D) 360
(E) 720
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?

(A) 600
(B) 720
(C) 1440
(D) 4320
(E) 4800
If these problems make your head spin, then you have found the right post!
The basics of counting
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.
Start with the restriction
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 nonrestricted 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 nonrestricted 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.
Practice problem explanations
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 permutation 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 BAG or GAB — just those two orders. This means the total number of configurations for these three children is 5*2 = 10.
Now, consider the nonrestricted 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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Hey,
Alternate solution for problem 4:
consider arrangement of A,B and C on seven seats. Imagine that seats for A and B are clubbed(we consider them as one seat), we need to choose 2 seats out of a total of 6 which can be done in 6C2 ways and A and B can be arranged in 2! ways. The remaining can be arranged in 4! ways.
Ans=6C2 * 2! * 4!=720
Never mind, I figured out my mistake after banging my head against it long enough.
5 * 4! + 4 * 4! + 3 * 4! …. is not the same as 5! * 4! Ooops!
Hey Chandler,
Sorry about the headbanging, but I’m happy to hear you figured it out! 😀
Have a great week,
Rita
Hi Mike,
I understand your logic on #3 and it is quite understandable, but when I first tackled #3 I used a different approach (similar to 1 and 2) but obtained an incorrect answer. Please correct my reasoning.
Assuming F is in the 6th seat, I have 5 possibilities for E:
E _ _ _ _ F
_ E _ _ _ F
_ _ E _ _ F
…
In each scenario, there are 4 permutations for the other letters, so there are 5 * 4! configurations when F is in the 6th seat and E is to the left of it.
Similarily, when F is in the 5th seat, I have 4 possibilities when E is to the left of F:
E _ _ _ F _
_ E _ _ F _
…
Thus in each scenario, I have 4 * 4! configurations when F is in the 5th seat.
After spotting a pattern, I figured that there are 5! * 4! configurations where E is to the left of F. This figure, of course is 2880. Please help me find out where I am going wrong. Thank you!
For Q3, if you want to make sure child E is somewhere to the left of child F, don’t you need a restriction on child F so that he/she doesn’t sit on the leftmost position of the row? So the calculation can’t simply be 6!/2, I think?
MaiLinh,
I’m happy to respond. 🙂 In my calculation, that was already taken into account. Here’s another way to say it: for each and every arrangement in which E is to the left of F, there’s a unique corresponding arrangement in which E is to the right of F, and we can pair those in a onetoone fashion; the two configurations are identical, except E & F switch places. Even if the configuration has F in the left most position {F, B, C, A, E, D}, it simply would correspond to one which had E in the leftmost position, {E, B, C, A, F, D}; the first does not satisfy the condition, and the second one does. For any configuration #1, we can switch E & F, and get configuration #2, and it must be true that exactly one of those two works and exactly one doesn’t. That’s precisely why 50% of all possible configurations work.
Does all this make sense?
Mike 🙂
Mike,
Can you please solve Q3 with the technique you used on the first two questions?
we cannot put F on the further left so, I have come up with 5x5x4!=600. what am I missing here?
Thanks
Kuntay
Hi Kuntay 🙂
Happy to help! Let’s look an alternative way to solve this problem, considering the relative position of E and F. Unlike in Q1 and Q2, E and F do not have to be next to each other in Q3. Rather E has to be to the left of F. The closest arrangements of the two will have E and F next to each other and the farthest arrangement will have the two seated at opposite ends. So, there can be 0, 1, 2, or 3 seats between the two children. The other 4 children can sit in any of the 4 seats left over, so there are 4! (=24) arrangements of the 4 children for each relative position of E and F. To determine the total number of arrangements, let’s figure out how many different ways we can seat E and F with 0, 1, 2, or 3 seats between them. We will then multiply each of these values by 4!
A. E F _ _ _ _
This situation is similar to that in Q1, and there are 5 ways we can have E and F seated next to each other.
B. E _ F _ _ _
There are 4 ways to have 1 seat between E and F.
C. E _ _ F _ _
There are 3 ways to have 2 seats between E and F.
D. E _ _ _ F _
There are 2 ways to have 3 seat between E and F.
E. E _ _ _ _ F
There is 1 way to have 4 seat between E and F.
Totals:
A. 5*4!
B. 4*4!
C. 3*4!
D. 2*4!
E. 1*4!
Sum = 15*4! = 360 arrangements
As you can see, this process involves many more steps than the solution provided. For that reason, I highly recommend you also understand how we can solve this problem using symmetry 🙂
I hope this helps!
I used Jones’s technique (which is also explained in detail within the Magoosh paid course) and it made these problems MUCH easier.
Dear Ali,
Very good. I’m glad you found a strategy that works for you. I would just caution you on reifying one strategy as “the best” — in the diverse problemsolving atmosphere of the GMAT, it’s best to have as many different solution methods at your disposal as possible. See:
https://magoosh.com/gmat/2013/multiplesolutionsingmatmath/
Thanks for sharing your experience, and the best of luck to you!
Mike 🙂
nice post Mike……
for Q 1 i m sharing another technique that could be helpful to many.
If we consider C and F one children , then we have 5 more,so total 6 children.
we can arrange 6!=720 ways
and between C and F we can arrange them 2!=2 ways
so total arrangement =720*2=1440
Dear Jones, yes, that’s perfectly correct. That conceptual leap of treating two children next to each other as a single child, at least for the first part of the problem, might be confusing for some folks, but for folks who get it, it’s a slick step. Thanks for sharing, and best of luck to you.
Mike 🙂