The post Does Order Matter? Combinations vs. the Fundamental Counting Principle on the GRE appeared first on Magoosh GRE Blog.

]]>Let’s look at a couple of examples to illustrate.

Suppose we have 5 people waiting for 3 seats. Now let’s say I want to count how many ways 3 people can be arranged in those seats. There are three tasks here–one for each seat.

*fill 1st seat, fill 2nd seat, fill 3rd seat*

We can write those as 3 blanks:

_____ , _____ , _____

With this done, we’re most of the way there. We just need to check if filling one seat with one person is the same as filling another seat with that same person.

Amy , _____ , _____

_____ , Amy , _____

Now, we know those aren’t the same, because that makes two different line-ups. Keep in mind which task each blank represents.** ***Amy in seat 1 is not the same as Amy in seat 2*** .** So this is

5 options , 4 options , 3 options

5*4*3 = 60.

If we were simply looking for a group of people–not paying attention to which person takes which chair, then we’d use a combination because the tasks aren’t inherently separate. There would be no significance of which blank represents which task. That might be asking “how many ways three people can be chosen for a committee from a group of five,” for example. In that case, we *can’t* draw the tasks as separate. Selecting the first person for the committee is the same as as selecting the second person. We could even choose all three at the same time, in one task–we don’t have to break it up into separate events. In that case, we have to use the combination formula as given in the our combination lesson videos.

5C3

5! / ([5-3]! *3!)

5! / ([2]! *3!)

5*4 / 2

10

Here’s another example. If you have 4 shirts, 4 pairs of pants, and 4 hats, and you choose one of each, is it a combination or does order matter? It’s tempting to say “order doesn’t matter”, because a t-shirt and jeans is the same as jeans and a t-shirt, but let’s try making those 3 tasks:

*pick a shirt, pick pants, pick a hat*

or

_____ , _____ , _____

Again, remember that each blank must represent one task alone, and we don’t move the blanks around. So if we say

red t-shirt, _____ , _____

That’s NOT the same as

_____ , _____ , red t-shirt

*A shirt in the first blank is not the same as a shirt in the last blank,* because the last blank is for choosing a *hat* and a shirt cannot be a hat. It’s not very fashionable at the moment, at least.

That means that we can use the fundamental counting principle here. First, we look at the shirt task. There are 4 shirts, so we have 4 possibilities. Next, and *separately, *we look at the pants. There are 4 pairs, so that’s another 4 possibilities for each shirt. So far, that’s 4*4. And finally, we have hats: there are another 4 hats, so that’s another 4 possibilities for each match up of shirt and pants. That’s 4*4*4.

So if we have *separate* blanks for each task, just fill in the blank with how many choices can be made in that specific step. Then multiply all the numbers for your answer.

On the other hand, if we were looking for ANY three pieces of clothing from the twelve total, we would use a combination formula. Again, this could be done in a single step–reach into a bag of clothes and grab three things. There’s no differentiation between them. This would give us 12C3, or 220.

Basically, this is all about drawing the blanks to represent the tasks. As you do so, ask if the item or person you’re picking can be moved around to different blanks without changing the situation. (Move the items–not the blanks!). If the item can be moved without changing the situation, it’s a combination–order doesn’t matter. If moving the item changes the situation (like in the line-up) or is impossible (like with the clothes), then we use the FCP.

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]]>Once in a while it is fun to take a concept in an official GRE question and run with it. The question below is inspired by a question from the 2^{nd} edition Official Guide. That is not to say it is that similar; I have only extracted a couple of concepts from that problem.

I’ve also made my question far more difficult. The first question was already difficult enough, so it is safe to say that a question of this difficulty would never appear on the GRE (though it could appear on the GMAT). Another of way expressing this: only attempt this question if you are already very strong in Quant. If not, just have fun with it, remembering that questions on the actual GRE will be easier.

1. A semicircle with area of xπ is marked by seven points equally spaced along the half arc of the semicircle, such that two of the seven points form the endpoints of the diameter. What is the probability of forming a triangle with an area less than x from the total number of triangles formed by combining two of the seven points and the center of the diameter?

- 4/5
- 6/7
- 17/21
- 19/21
- 31/35

Again this is a very difficult question, one that requires many steps, and one that has numerous twists. The question is without a doubt expansive, covering in its sweep combinatorics, probability, basic geometry, advanced triangle theory, counting properties and much more. So hang on for the ride!

In order to deal with this whole area of the triangle business, we first have to figure out the area of the circle. That way, we can also derive the radius, a number that be instrumental in helping us in Step #3.

First though we find the area of the circle, which is twice that of the semi-circle . To solve for the radius, we use .

So we have the radius. But how does that help in figuring out the resulting area of any of the inscribed triangles? Well, remember the seven points? If two points are different sides of the diameter and the other five points are evenly spaced along the arc, then the semicircle is broken into equal sixths. Therefore the central angle of each will be 30 degrees (the arc of a semicircle corresponds to 180 degrees).

** **

Using the information above, we can start to play around with triangles, keeping in mind that any two points along the arc will form radii with the center of the diameter (remember that the original question asked us to make triangles using the middle of the diameter).

A good idea is to come up with a triangle that both meets the criteria and is easy to solve area-wise. That way you can see how close you are to an area of x. For instance, you could assume that the triangle is an equilateral, the two points on the arc a distance of 60 degrees away (and thus the central angle will be 60).

A radius of will form the sides of the equilateral. Using the formula for the area of the equilateral triangle, , where s equals the side of the triangle, or in this case the radius, we get . Because equals approximately 1.7, we can see that such an equilateral triangle has an area less than x.

But don’t give up just yet. In trying to increase the area you should keep in mind that skinnier triangles, say with points on the opposite side of the semicircle, will not have an area greater than that of an equilateral. So we want to increase the angle but not too much.

The next most sensible triangle—that is one in which we can use simple formulas to determine the area—is a 45:45:90 triangle, which means that the two points on the semicircle will have to be 90 degrees apart (that way the central angle will be 90 degrees. And with two equal sides (remember all radii are equal), we know have a 45:45:90 triangle with sides of . Using the formula for the area of a triangle, we get .

An important insight: if we bring the points together, we end up with an equilateral, which we know to have a smaller area. Therefore, moving the points on the arc together results in a triangle with less area. Moving the points apart makes the triangle skinnier, something that also results in a triangle with less area.

Therefore, the isosceles right triangle represents the largest possible triangle, area-wise, that you can inscribe given the conditions of the problem (pretty advanced stuff, right?).

Voila! Finally, we have it. The only possible triangle to yield x will be an isosceles right triangle, or a 45-45-90, with a central angle of 90. If we find the number of these triangles and subtract that from 1, we have the probability of choosing a triangle with an area less than 1.

The total number of isosceles right triangle is four, which we can find by choosing points along the circle that are 90 arc degrees apart.

Next we want to find the total number of triangles. Because one point is already fixed (the center of the diameter), we have to determine how many triangles can be formed by using two of the seven points. The ordering of the vertices is not important. Meaning that a triangle with points A, B, and C, is the same as triangle with points B, C, A. Therefore we use 7C2 = 21.

Just when you thought we were finally done with this monstrosity of the problem, there is one final, and subtle twist. We have to discount one of these twenty-one triangles because it is actually not a triangle. If we choose both end points of the diameter and the center of the diameter, we have a straight line (this line is the diameter of the semicircle).

Therefore, there are twenty triangles, sixteen of which have an area less than x. so the answer is (A) 4/5.

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]]>The 2^{nd} edition Official Guide has just been released. At the back of the book is a new test (GRE Practice Test 2). In general, the concept break down is not surprising, if you look at the concepts that pop up in Practice Test 1 (which is the same as the practice test in the first edition of the Official Guide).

Word problems abound. There is a fair amount of algebra. Even a few combinatorics problems rear their fearsome heads. And if you’re wondering how many statistics questions are on the GRE, a topic often given scant attention in many prep books – it can pop up in spades. Most interestingly, coordinate geometry only shows up once, and there is not a single rate problem.

The information below pertains to the 50 quantitative questions in Practice Test 2. I have lumped QC and Problem Solving questions together. I have not assigned only one category per question, but have ascribed multiple categories to the same question. For instance, a word problem can contain any number of concepts. To only call it a word problem is a vague. After all ten Word Problems on combinations is very different from ten word problems dealing with percents.

The breakdown of concepts detailed below will likely be very similar to what you see test day. That said you would probably see more than one coordinate geometry question and would probably see a rate question. And just because a concept did not show up on the test, does not mean you will not see that concept test day. Interest problems, both simple and compound, can very well show up.

What is the take away from all this? If you are weak in a certain area, say word problems that deal exclusively with adding a series of numbers, do not fret about this. You may only see one problem, if any, dealing with such a concept. On the other hand, if you are only somewhat comfortable with Statistics it is a good idea to strengthen this area, as you are far more likely to see it test day.

Word Problems (15)

Algebra (9)

Geometry (7)

Statistics (6)

Exponents (5)

Rates (0)

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]]>The post The Difficulty of Context: Combinations and Permutations Questions appeared first on Magoosh GRE Blog.

]]>Over the last couple of months, I have covered a fair amount of GRE combinations and permutations questions on the blog. I’ve gone from the very basic to the very difficult (so difficult that they are more ‘fun problem’ than practical practice).

Often, when you are stumped on a problem, it is not that the math is difficult, but that the context may be unfamiliar.

Combinations/permutations is an excellent example of this phenomenon. If you’ve been watching my videos on this topic, you should be very comfortable with the math. But that doesn’t mean that on test day you will be able to answer a combinations/permutations question correctly. To demonstrate, I will give you the following questions:

1. How many groups of three can you form if you choose from a total of seven?

2. How many ways can you sit seven students in three chairs (one chair per student)?

3. How many different triangles can you form by connecting the vertices of a heptagon?

4. A doctor is to administer pyroxeline, damalnethamide, and cyrotinin to his patients, a treatment plan that is to take place over the course of the week. If he can administer each drug once and cannot give more than one drug a day, how many unique schedules of medications can he create, if the order in which he administers the drugs is important?

While these questions may seem to be in wildly divergent contexts, each question can be answered with either 7C3 or 7P3. Indeed, I could have actually included a dozen more examples, all of which have very different contexts. The key – besides lots of practice – is learning the best way to think about how to solve problems, and not to rely on the problems surface appearance to inform your approach.

For instance, do not think that all polygon questions that talk about joining vertices call for a similar approach to the one above. Likewise, if a question talks about a doctor administering different drugs, don’t think, *oh yeah, I remember that on Chris’ blog post, and he used the permutations formula, not the combinations formula.* Your goal should be to understand why that specific problem employed permutations, and not combinations.

The test writers are very aware of the vast number of possible contexts. And, while the math won’t change too much, they will often wrap familiar math concepts in misleading guises. Cracking the problem will be more a matter of choosing the correct approach than applying a given formula.

That said, I will leave you with a combinations/permutations question I encountered in real life…

**A Quick Anecdote**

My wife and I were set to board a plane. We knew we were in the same row, but my wife had forgotten whether the tickets were for adjacent seats. (There were very few seats left so my wife only remembered purchasing seats that were at least in the same row).

On this flight, the seats were arranged six in row (A – F), with the aisle splitting the seats into A-B-C and D-E-F. Knowing that I spend my day writing math problems, my wife asked me, “So, what is the probability we will be sitting next to each other?”

I thought about it for a moment, imagining us seated in different positions. I was about to begin the laborious process of mentally arranging us in all the different possible positions, but luckily stopped this brute force method and instead thought of the problem mathematically:

___ ___ ___ ___ ___ ___

My wife and I can be thought of as A and A. I then asked myself how many ways I could place A and A in the dashes above. Because A is redundant (as long as my wife and I are sitting next to each other, it doesn’t matter who is sitting on which side), I didn’t have to worry about order. I was essentially asking how many ways I could choose two dashes from six dashes. In formula-speak, this translates to 6C2 = 15.

Looking at the dashes above, I can only place A and A in four different positions, so that they are next to each other (remember, the space between the third and fourth dash is the aisle). Therefore, to find the probability of an event (my wife and I sitting next to each other), I divide the total number of possibilities (15 different ways that two people can sit next to each other) by the desired outcome – my wife and I seated next to each other: 4 possibilities.

I turned to my wife, eager to share my mathematical epiphany, “The chances of us sitting together was 4/15 or a little more than 25%. So, we are likely going to have to ask somebody to switch.”

Before I’d even gotten my last syllable out, my wife stopped me: “Actually, we are sitting together”. Holding both tickets in her hand, she added, “So much for your math.”

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]]>*These challenge questions are great for extra practice. However, be warned: while I provide the answers, I don’t provide the explanations. If you’d like to try practice problems with explanations, check out Magoosh’s 15 GRE math practice problems and my biweekly Brain Twisters!*

Over the last few weeks, I have gone through GRE combinations and permutations. My goal is to make this formidable concept less intimidating. If you’ve watched the videos and feel comfortable with the math, I would say continue practicing but do not worry too much about this problem type. You will probably see one, though you will unlikely see two, on the actual test. So your time will better be spent on other concepts, such as work rate, sets, series, and coordinate geometry. (I will be covering these concepts soon!). Unless, that is, you are aiming for the 160+ category/the top 90% (it’s great that I can finally speak of the Revised GRE in terms of the Revised GRE scoring). The combinations and permutations I covered, while tough, may not be as difficult as a question you may see on the test, if ETS decides to make one of its most difficult questions a combinations and permutations one. Thus I have decided to devise some pretty diabolical questions. In fact, the last two are so convoluted and tricky that they wouldn’t even show up on the test. Again, this is only for those who are already comfortable with combinations and permutations, and who are going for the near perfect score. Otherwise, this could be a waste of your time (of course the less practical side of me thinks that this level of problems is fun!).

**Warming Up** An equal number of juniors and seniors are trying out for six spots on the university debating team. If the team must consist of at least four seniors, then how many different possible debating teams can result if five juniors try out?

(A) 50 (B) 55 (C) 75 (D) 100 (E) 250

**Difficulty Level: Hard** A bibliophile plans to put a total of seven books on her marble shelf. She can choose these seven books from a mixture of works from Antiquity and works on Post-modernism, of which there are seven each. If the shelf must contain at least four works from Antiquity, and one on Post-Modernism, then how many ways can he select seven books to go on the shelf?

(A) 441 (B) 1225 (C) 1666 (D) 1715 (E) 1820

**Difficulty Level: Are you kidding me?!?** An artist is planning on mixing together any number of different colors from her palette. A mixture results as long as the artist combines at least two colors. If the number of possible mixtures is less than 500, what is the greatest number of colors the artist could have in her palette? (A) 8 (B) 9 (C) 11 (D) 12 (E) 13

**Check out the answers:** 1. B 2. D 3. A

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]]>As a continuation of my introduction to Combinations and Permutations, here are some practice problems to test the strategies I demonstrated. Also, I am trying out a new format. Instead of giving you practice problems and then writing out a half-page description for each one, I am simply going to give you the questions. But don’t worry – I am not leaving you in the lurch. For explanations, I am solving each of the questions – using my special approach, of course – in the video below.

**Questions:**

1. A committee is composed of a president, a vice president, and a treasurer. If six people are trying out for the three positions, how many different committees result?

(A) 20 (B) 40 (C) 60 (D) 105 (E) 120

2. A committee of three is to be chosen from six. How many unique committees result?

(A) 20 (B) 40 (C) 60 (D) 105 (E) 120

3. A committee is composed of a president, a vice president, and a treasurer. If five people are running for president, six people are running for vice president, and three are running for treasurer, how many unique committees result?

(A) 15 (B) 45 (C) 75 (D) 90 (E) 120

4. A jousting tournament requires that a team consist of two knights and two squires. The Merry Band is forming a team from five knights and three squires. How many different lineups can The Merry Band field?

(A) 10 (B) 13 (C) 15 (D) 30 (E) 120

5. A septet, a group composed of seven players, is made up of four strings and three woodwind instruments. If seven students try out for strings and seven different students try out for woodwinds, how many unique septets can result?

(A) 35 (B) 70 (C) 210 (D) 420 (E) 1225

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]]>** **Whenever I see a GRE resource label its counting section as “Combinations and Permutations,” a small part of me dies a little. Okay, that’s an exaggeration, but I am concerned about the misleading message that this sort of title conveys. To me, it suggests that counting questions can be solved using either permutations or combinations, when this is not the case at all.

The truth of the matter is that true permutation questions are exceedingly rare on the GRE.

Now, for those who are unfamiliar with permutations, a permutation is an arrangement of a subset of items in a set. To be more specific:

**If we have n unique objects, then we can arrange r of those objects inways, whereequals some formula that I still haven’t memorized even though I took several combinatorics courses in university, and I taught counting methods to high school students for 7 years.**

Now, it’s not that the permutation formula is too complicated to remember; it’s just that it’s unnecessary to memorize such a formula for the GRE. In my humble opinion, the permutation formula has no place in a GRE resource (even though the Official Guide covers it).

Here’s an example of a true permutation question:

**Using the letters of the alphabet, how many different 3-letter words can be created if repeated letters are not permitted?**

Here, we have a set of 26 letters in the alphabet, and we want to determine the number of ways we can arrange 3 of those letters. So, if we still feel compelled to use permutations to the answer the question (despite my public denouncement of permutations :-)), the answer would be at which point you would have to evaluate this.

Of course, you’re not going to memorize the permutation formula, because you’re going accept my premise that true permutation questions are exceedingly rare on the GRE. For the doubters out there, let’s consult the Official Guide to the GRE Revised General Test. In the Guide, there are 7 counting questions altogether. Of these 7 questions, not one is a true permutation question (although some will argue that question #6 on page 297 is a permutation question, albeit a very boring one that can be solved using an easier approach).

So, given the rarity of permutation questions, it’s dangerous to approach counting questions with the notion that all you need to do is determine whether you’re dealing with a combination or a permutation, and then apply one of two formulas. If you do this, you will inevitably conclude that a question is a permutation question when it isn’t. Notice how easy it is to turn a permutation question into a non-permutation question by simply changing a word or two. For example, see what happens when we change our original question to read:

**Using the letters of the alphabet, how many different 3-letter words can be created if repeated letters are permitted?**

By allowing repeated letters, the question is no longer a permutation question, which meansis not the solution (the solution is , which we’ll cover in future posts).

The truth is that we don’t need the permutation formula to answer any counting question on the GRE (including question #6 on page 297 of the Official Guide). Instead, we can use the Fundamental Counting Principle (FCP). The FCP is easy to use and it can be used to solve the majority of counting questions on the GRE.

So, my approach with all counting questions is as follows:

- First, determine whether or not the question can be answered using the FCP
- If the question can’t be answered using the FCP, it can probably be solved using combinations (or a combination of combinations, and the FCP)

In the next series of posts, I’ll explore how we can use the FCP to solve most counting questions, and how we can use combinations to solve the rest.

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]]>The post Prime Numbers and Probability – GRE Math Problem appeared first on Magoosh GRE Blog.

]]>Oftentimes, math problems will require that you know more than one concept. Sometimes, you’ll have to know the difference between an integer and a number in order to get the question right. Below, I’ve combined prime numbers with probability, two subjects already covered in this blog.

The question below isn’t easy and actually takes a little bit of work. Let’s see if you can crack it.

*What is the probability that the sum of two rolled dice will equal a prime number?*

*(A) **1/3*

*(B) **5/36*

*(C) **2/9*

*(D) **13/36*

*(E) **5/12*

First let’s list the prime numbers that pertain to this problem.

The pertinent prime numbers are 2, 3, 5, 7, and 11. Notice I stopped at 11. Why? Well, the greatest number you can roll on two six-sided dice is 12.

Next, we have to remember this is a probability question. Therefore, we have to divide the number of total outcomes by the number of desired outcomes. First, let’s find the number of desired outcomes. To do this, we have to make sure that each desired outcome conforms to the problem, i.e. how many different ways can we sum two dice to get a prime.

Let’s start with 2. There is only one way to roll a 2, and that is with a 1 and a 1. Therefore, we have one desired outcome.

What about the number 3? Well, we can get 2,1 and 1,2, or two possible ways.

Mind you, to do this problem you will have to make sure you write down each outcome. Do not try to do this in your head, for you’ll most likely get the problem wrong and induce dizziness.

Next we have the number 5, which we can get by rolling the following combinations:

1,4

4,1

2,3

3,2

We add these four possible outcomes to the prior three, giving us a total of 7.

Next, we look to see which numbers sum to 7 and find a total of six possibilities.

1,6

2,5

3,4

4,3

5,2

6,1

Our total is now 13.

Finally, don’t forget 11 as final prime number.

6,5

5,6

We add these two possibilities to the 13 possibilities, giving us a total of 15.

Finally, we want to find the total number of outcomes. Remember, we divided the total outcomes by the total number of desired outcomes, which we just found out was 15.

The total number of outcomes equals the number of different ways we can roll two six-sided dice. We find this number by multiplying 6 x 6. The logic is there are six sides to each die, so for each number on one die you can pair with six different numbers on the other die.

Therefore, the probability of rolling a prime number on two dice is 15/36, which reduces to 5/12 (E).

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]]>Over the years, as the math section has become more difficult, permutations and combinations are popping up more often. At the same time, students are also becoming more adept at handling these kinds of problems (I’d hypothesize that more practice problems are available.) As a result, permutations and combinations problems are not only more common but also more difficult.

What I’ve done is come up with the kind of permutation/combination problems, you can expect to see at the 700+ level. These questions are not easy, so do not get discouraged.

In fact, the first person to get all five problems correct, before I post my video reply at the end of the week, will win a one-month subscription to Magoosh.

Good luck!

1. A committee of three must be formed from 5 women and 5 men. What is the probability that the committee will be exclusive to one gender?

(A) 1/60

(B) 1/120

(C) 1/8

(D) 1/6

(E) 1/3

2. A three-letter code is formed using the letters A-L, such that no letter is used more than once. What is the probability that the code will have a string of three consecutive letters (e.g. A-B-C, F-E-D)?

(A) 1/55

(B) 1/66

(C) 2/17

(D) 1/110

(E) 2/55

3. A homework assignment calls for students to write 5 sentences using a total of 10 vocabulary words. If each sentence must use two words and no words can be used more than once, then how many different ways can a student select the words?

(A) 10!/5!

(B) 10!/32

(C) 5! x 5!

(D) 2! x 5!

(E) 10!

4. Team S is to comprise of n debaters chosen from x people? Team R is comprised of n + 1 debaters chosen from x+1 people.

Column A

Number of unique team S

Column B

Number of unique Team R

5. A lunar mission is made up of x astronauts and is formed from a total of 12 astronauts. A day before the launch the commander of the program decides to add p astronauts to the mission. If the total number of possible lunar missions remain unchanged after the commander’s decision, then which of the following cannot be the value of p?

(A) x

(B) x + 3

(C) 3

(D) 6

(E) 8

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]]>On a recent post I went over factorials. Now let’s try a word problem that actually uses the factorial sign.

*Five students must sit in 5 chairs. How many different ways can they sit?*

(A) 30

(B) 60

(C) 4!

(D) 5!

(E) 6!

In this problem we have a set number of people. Our job is to find how many different ways to arrange them. This is what we call a permutations problem.

Two features of a permutations problem are as follows: the number of members in the group is fixed, and we are looking for the total possible ways to arrange those members.

To find the answer, we simply put the number in the group followed by a factorial sign: 5! = 5 x 4 x 3 x 2 x 1 = 120. In the question, you don’t have to write out the factorial and can directly answer (D).

Now what if we weren’t too concerned about who was sitting in a particular chair. Instead, we wanted to choose 5 students from a larger group—let’s say 8. How many different ways could we select 5 from a group of 8 to sit in the five chairs?

Notice that I am not using the word arrange, the way I did with a permutation problem. With a combinations problem the order of the people in the group (in this problem, the order in which they are sitting) does not matter. So if Steve and Mary are in the group of students to be seated, it doesn’t matter which chair they are sitting in.

The formula for a combinations problem is n!/(n – p)! p!, where n is the total number you are choosing from and p is the number you are choosing. In this problem, n = 8 (there are 8 members in the group) and p = 5 (we are choosing 5 students to sit in the five chairs.)

Plugging in these numbers we get:

8!/3!5! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1/(5 x 4 x 3 x 2 x 1) (3 x 2 x 1)

Notice that the 5 x 4 x 3 x 2 x 1 cancels out from both the numerator and the denominator. That leaves us with 8 x 7 x 6/3 x 2 x 1 = 56. That is, there are 56 different ways to organize the group.

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