1) A librarian has 4 identical copies of *Hamlet*, 3 identical copies of *Macbeth*, 2 identical copies of *Romeo and Juliet*, and one copy of *Midsummer’s Night Dream*. In how many distinct arrangements can these ten books be put in order on a shelf?

(A) 720

(B) 1,512

(C) 2,520

(D) 6,400

(E) 12,600

2) A librarian has a set of ten different books, including four different books about Abraham Lincoln. The librarian wants to put the ten books on a shelf with the four Lincoln books next to each other, somewhere on the shelf among the other six different books. How many different arrangements of the ten books are possible?

3) Over the course of a full-day seminar, 66 students must give short presentations. Most of the presentations are independent, but Sam’s presentation refers to something in Ruth’s, so Ruth must go before Sam; similarly, Matt’s presentation refers to something in Ruth’s and something in Sam’s, so Matt’s must come after both of those. These three presentations need not be consecutive with each other: whatever the order is, Ruth’s must come before the other two but not necessarily first of the 66, and sometime later, Matt must come after the other two but not necessarily last of all. The other 63 participants can present in any order. How many orders obey these constraints?

(A) 22!

(B) 63!

(C) 63!/3!

(D) 66!/3

(E) 66!/3!

4) A bag contains ten marbles of the same size: 3 are identical green marbles, 2 are identical red marbles, and the other 5 are five distinct colors. If 5 marbles are selected at random, how many distinct combinations of five marbles could be drawn?

(A) 41

(B) 51

(C) 62

(D) 72

(E) 82

5) In a bag, there are five 6-sided dice (numbered 1 to 6), three 12-sided dice (numbered 1 to 12), and two 20-sided dice (numbered 1 to 20). If four of these dice are selected at random from the bag, and then the four are rolled and we find the sum of numbers showing on the four dice, how many different possible totals are there for this sum?

(A) 61

(B) 106

(C) 424

(D) 840

(E) 960

6) A chessboard is an 8×8 array of identically sized squares. Each square has a particular designation, depending on its row and column. An L-shaped card, exactly the size of four squares on the chessboard, is laid on the chessboard as shown, covering exactly four squares. This L-shaped card can be moved around, rotated, and even picked up and turned over to give the mirror-image of an L. In how many different ways can this L-shaped card cover exactly four squares on the chessboard?

(A) 256

(B) 336

(C) 424

(D) 512

(E) 672

7) A shipping company has four empty trucks that will head out in the morning, all four to the same destination. The clerk has four different boxes to ship to that same destination. All four boxes could go on any one of the trucks, or the boxes could be split up into any groupings and given to the trucks in any combinations (ie. two to one truck, one to another, and one to another). In how many different ways could the boxes be put on the four trucks?

(A) 16

(B) 64

(C) 256

(D) 576

(E) 4096

8) Suppose we have six marbles: 3 blue marbles, 2 red marbles, and one green marble. Suppose we are going to put them into three cups: a black cup, a white cup, and a purple cup. We could put all six in any cup and leave two cups empty; or we could put marbles in two cups and leave one cup empty; or we could put some marbles in each of the three cups. How many combinations are possible?

(A) 90

(B) 180

(C) 360

(D) 540

(E) 720

9) Suppose we have six marbles: 3 blue marbles, 2 red marbles, and one green marble. Suppose we are going to put them into three cups: a black cup, a white cup, and a purple cup. The only restriction is that the two red marbles can’t be in the same cup. We could put as many as five (all except one of the reds) in any cup. We could leave one cup empty, or put some in each of the three cups. All combinations are allowed that don’t involve the two red marbles in the same cup. How many combinations are possible?

(A) 90

(B) 180

(C) 360

(D) 540

(E) 720

10) In a certain mathematical activity, we start with seven cards, each with a different prime number written on it. These seven cards are randomly put into three boxes in the following way: one box must get four cards, one must get two, and one gets only one. Then, for each box, we find the product of all the cards in the box, and that’s the “number” of the box. Then, we put those three numbers in order, from lowest to highest, and that is our set. How many different sets can be created from this process?

(A) 35

(B) 105

(C) 210

(D) 420

(E) 630

11) In a certain mathematical activity, we have five cards with five different prime numbers on them. We will distribute these five cards among three envelope: all could go in any envelope, or they could be broken up in any way among the envelopes. Then in each envelop, we find the product of all the cards in that envelope: that is the “number” of the envelope. An envelope containing no cards has the number 1. We then put the three envelope numbers in order, from lowest to highest, and that is our set. How many different sets can be produced by this process?

(A) 41

(B) 89

(C) 125

(D) 243

(E) 512

12) In the diagram below, points A and B are on opposite corners of a lattice consisting of 12 segments. A “true path” from A to B is a path on which no segment is traversed more than once. How many “true paths” from A to B are there?

(A) 6

(B) 10

(C) 12

(D) 14

(E) 18

Solutions will appear at the end of this article.

## Counting problems on the GMAT

The following blogs cover some of the basic ideas of counting problems:

1) The Fundamental Counting Principle

2) Permutations and Combinations

4) Difficult Counting Problems

The first three just cover the basic rules. The fourth, in addition to four hard practice problems, has a discussion of some of the reasons why counting is such a challenging topic. I would refer you to that discussion, and would simply add: if you found the problems above challenging, then read the solutions very careful. The hardest part of a counting problem is often how to frame the problem, how to parse the situation, how to imagine it in do-able stages: once these choices are made, the application of the rules is often quite straightforward. How do you learn how to frame a counting problem? Well, first of all, pay attention to how solutions do this, starting with the solutions below. For more on right brain skills, see this post.

## Summary

If the related blogs and the solutions here gave you some new insights, please let us know about this in the comments section!

1) If the ten books were all different, the arrangements would be (10!), a very big number. Because we have repeats of identical copies, not all 10! arrangements will be unique. For a subset of k identical members, those k members could be interchanged in k! orders, and the resulting arrangements would be the same, so we have to divide by k! to remove repetitions. In this example, we need to divide 10! by 4! and 3! and 2!:

Answer = **(E)**

2) Start out by thinking of the four books about Lincoln as one big book. This is one book, and we have six others, so we start by figuring out the arrangements of these seven books. That’s 7!

Now, for each one of those 7! arrangements, we can put the four Lincoln books in four different orders. Thus, for each 7! arrangements of the Lincoln slot with the six other books, we have 4! variants because of the order of the Lincoln books. That’s a total of

N = (7!)(4!)

Answer = **(C)**

3) If the constraints with these three individuals didn’t matter, there would be 66! arrangements. Any one arrangement will have those three in some particular order. If we kept the other 63 participants in the same places, and just re-arranged these three, there would be 3! = 6 possibilities:

We could group all (66!) arrangements into groups of 6 like this, and in each case, only 1 of the 6 would have the correct order for these three individuals. Thus, we need to divide (66!) by 3! = 6.

N = (66!)/(3!)

Answer = **(E)**

4) Here, we have to do a combination of listing and calculating. We have to think about several individual scenarios. First, the “three green” scenarios. By “other color”, I mean the colors other than green and red; there are five of these.

Scenario #1: 3 green, 2 red = 1 possibility

Scenario #2: 3 green, 1 red, 1 other color = 5 possibilities

Scenario #3: 3 green, 2 other colors = 5C2 = 10 possibilities

Now, the “two green” scenarios:

Scenario #4: 2 green, 2 red, 1 other color = 5 possibilities

Scenario #5: 2 green, 1 red, 2 other colors = 5C2 = 10 possibilities

Scenario #6: 2 green, 3 other colors = 5C3 = 10 possibilities

Now, the “one green” scenarios:

Scenario #7: 1 green, 2 red, 2 other colors = 5C2 = 10 possibilities

Scenario #8: 1 green, 1 red, 3 other colors = 5C3 = 10 possibilities

Scenario #9: 1 green, 4 other colors = 5C4 = 5 possibilities

Now, the “no green” scenarios:

Scenario #10: 2 red, 3 other colors = 5C3 = 10 possibilities

Scenario #11: 1 red, 4 other colors = 5C4 = 5 possibilities

Scenario #12: 5 other colors = 1 possibility

Now add these.

(10 + 10 + 10 + 10 + 10 + 10) + (5 + 5 + 5 + 5) + 1 + 1 = 60 + 20 + 2 = 82

Answer = **(E)**

5) This is not really a counting question, in that it doesn’t involves any of the standard counting techniques. We just have think about this logically. No matter what four dice we pick, the lowest roll we could get is a “1” on each of the four dice, for a total of 4. We could get any integer value from 4 up to the highest value. The highest value would occur if we picked the two 20-sided dice and two of the 12-sided dice, and got the highest value on each die: 20 + 20 + 12 + 12 = 64. We could get any integer from 4 to 64, inclusive. For this, we simply need inclusive counting. 64 – 4 + 1 = 61.

Answer = **(A)**

6) Consider the L in its current orientation, and start with it in the lower left corner.

We can move this up so that the top square is in any of the five empty squares above it: those, plus this original, is 6 positions. Now, we can move any of these six to the right one space, then again, then again, until we have moved it to the sixth new space, when the right side of the L will come up flush against the rightmost boundary. That’s seven total columns, each with 6 positions, for a total of 42 while it is in this orientation.

Clearly, we can rotate by 90° clockwise, and we would have 42 new positions for that orientation. Then we can rotate by 90° clockwise again, and again. Four orientations, each with 42 positions, for 42*4 = 168 positions.

All of this is for the “forward L.” Now, if we pick up the card, and put it down flipped over, to get a “mirror image L,” this again will have 42 positions in each of 4 rotated orientations, for another 168 position.

The total is 2*168 = 336

Answer = **(B)**

7) Where we put one box has absolutely no bearing on where we put any of the other boxes. The placement of the four boxes is completely independent of one another. For each box, we have four choices.

N = 4*4*4*4 = 16*16 = 256

Answer = **(C)**

8) First, consider possibilities the 3 blue marbles.

Blue Case I: all three in one cup = 3 possibilities

Blue Case II: two in one cup, one in another è three choices for the cup with two, then two choices for the cup with one = 6 possibilities

Blue Case III: one in each cup = 1 possibility

total = 3 + 6 + 1 = 10 possibilities for the blue marbles

Now, the 2 red marbles.

Red Case I: two in one cup = 3 possibilities

Red Case II: one in cup, one in another = 3 possibilities for which cup is empty

total = 6 possibilities for the red marbles

Now, the green marble can simply go in one of the three cups = 3 possibilities.

By the Fundamental Counting Principle, multiple the blue & red & green possibilities:

N = 10*6*3 = 180

Answer = **(B)**

9) First, consider possibilities the 3 blue marbles.

Blue Case I: all three in one cup = 3 possibilities

Blue Case II: two in one cup, one in another è three choices for the cup with two, then two choices for the cup with one = 6 possibilities

Blue Case III: one in each cup = 1 possibility

total = 3 + 6 + 1 = 10 possibilities for the blue marbles

Now, the 2 red marbles.

Red Case I: two in one cup = FORBIDDEN!

Red Case II: one in cup, one in another = 3 possibilities for which cup is empty

total = 3 legal possibilities for the red marbles

Now, the green marble can simply go in one of the three cups = 3 possibilities.

By the Fundamental Counting Principle, multiple the blue & red & green possibilities:

N = 10*3*3 = 90

Answer = **(A)**

10) Part of the logic of the problem involves recognizing that every unique combination of prime numbers produces a unique product. There is no way that two different groupings cards will produce the same set of numbers.

First of all, the cards to go into the cup holding 4 cards.

;

Once we have placed 4 card in the first cup, we have three cards left, which means we have three choices of a single card to put into the third cup. Three choices. Once we place that, the remaining two cards must go into the second cup: no choice there.

N = 35*3 = 105

Answer = **(B)**

11) We have to consider different groupings. First, all the cards together.

Case I: all five cards in one envelope (2 envelopes empty) = one possibility

Now, two envelopes used, one empty.

Case II: four in one, one in another, one empty = five choices for the one by itself = 5 possibilities

Case III: three in one, two in another, one empty = 5C3 = 10 possibilities

Now, no empty envelopes, all three used:

Case IV: 3-1-1 split = 5C3 = 10 possibilities

Case V: 2-2-1 split = 5C2 = 10 for the first pair, then 3 possibilities for the single one, leaving the other two for the pair. This would be 10*3 = 30, but that double-counts the two pairs, so we need to divide by two. 15 possibilities.

N = 1 + 5 + 10 + 10 + 15 = 41

Answer = **(A)**

12) First of all, consider the minimum paths, the paths of just four segments from A to B. Each one of them will consist of, in some order, two horizontal segments and two vertical segments. How many of these are there? Well, how many ways can we distribute two horizontal segments among four slots? 4C2 = 6. We put the horizontal segments in two slots, and then the two vertical segments must go in the remaining slots. There are six minimum slots.

Now, we have to consider paths that take more than four segments to get from A to B. Consider the possibilities that begin: (down) then (right) then (up). The two possibilities are

- a) (down)(right)(up)(right)(down)(down)
- b) (down)(right)(up)(right)(down)(left)(down)(right)

Now, consider the possibilities that begin (down)(down) then (right) then (up). The two possibilities are:

- c) (down)(down)(right)(up)(right)(down)
- d) (down)(down)(right)(up)(up)(right)(down)(down)

Those are all the possibilities for path that start (down). We also could start with a first step of (right), but those possibilities are just the reflections of these over an imaginary mirror line from A to B. There are four path that start (down), and four mirror images that start (right). That’s 8 longer-than-necessary paths, to add to the 6 minimum paths, for a total of 14 paths.

Answer = **(D)**

Mike,

I am just a bit confused about number 11. Why is it that we are adding all of the different possibilities instead of multiplying them together like in #10?

Thank you

Hi Rebecca,

We add rather than multiply in the case of #11 because it is not possible to have, for example, Case I and Case II occur simultaneously. That is what we are calculating if we multiply the probabilities we calculate separately (and then add). We want to identify all the possible cases (Case I through V) and then, because they are mutually exclusive, we must add them together.

I hope that helps! 🙂

HI,

I am a bit confuse with Q3, if oonly one arrangemennt out of the 6 is possible, why you still use the 3! , and not maybe just 1/6 ?

Thanks 😀

Hi Lucy,

We use 3! because that is the standard formula approach to problems like this. In this case, 3! is 6 which means that dividing by 3! or multiplying by 1/6 results in the same thing!

If we had to divide by something larger, like 5!, it is easier to keep track of our notation using factorials rather than trying to make the corresponding fraction. 🙂

Dear Mike,

I would be glad if you could clarify the answer to 12th question. Is the answer 14 or 6.

Dear Ashit,

Thank you very much for pointing out that inconsistency. That practice question had the wrong OA posted. I just corrected the mistake. Thank you very much!

Mike 🙂

Dear Mike,

could you, please, clarify the difference between Q7 and Q8. Why is it legit to use 4^4 in Q7, but it is not meaningful to calculate 6^3 in Q7? When, in general, should we use x^y approach?

WBR,

Alex

Alex,

I’m happy to respond. 🙂 One of the hard things about counting problems is that you always have to be careful not to count redundancies. In #7, all four boxes are different: that’s very clearly stated. If we switch the place of any two boxes, we will have a meaningfully different way of shipping the boxes. No redundancy is possible because all the items are different.

In #8, we have identical items. Right there, we have a world of difference!! If we had six completely different marbles, and were putting them in three cups, that indeed would be 6^3 ways. The problem is, once we have sets of identical marbles, then we could switch, say, two of the blue marbles and we would have a second arrangement that was identical to the first. This would be a redundancy. That 6^3 would include a whole donkey-cart load of redundancies, from all the possible interchanges of identical red or blue marbles. That’s precisely why the quick exponent approach is doomed from the outset. What I showed in the solutions, as always, is about a efficient solution as the scenario would allow. I would have no motivation to keep an efficient solution a secret from the readers of this blog! 🙂

If you want an easy formula-based approach to counting, you will always be frustrated. Quick easy formulas are seldom applicable, and each new counting scenario forces you to engage with the specifics of that situation. I would suggest that you read my thoughts in this blog:

https://magoosh.com/gmat/2013/difficult-gmat-counting-problems/

Does all this make sense?

Mike 🙂

Mike,

I just wanted to say thanks for the excellent and clear explanations of the many Quant concepts. It has helped my test preparation plan enormously.

Dear Rob,

My friend, you are more than welcome! 🙂 I am very glad you found these helpful! If you can handle problems of this calibre, then I would say you are in great shape for test day! Best of luck, my friend!

Mike 🙂

Hi Mike, appreciate your contribution on this blog and the numerous tips posted otherwise :-). Over here for problems 4,8 and 9, can the standard combination formula be applied in a more straightforward manner rather than the approach above which although simple is a little long.

Dear Mayank,

The short answer is “No.”

My friend, think about your question. Think about my motivation in writing this blog. I have absolutely no interest in showing a longer tedious solution if a quick and elegant one is possible. Elegance in mathematical solutions is exquisitely beautiful, and I would never pass it up if it were possible. Therefore, if what I show in the solutions is something long and complicated, you can be quite sure something more elegant was not available.

In fact, you don’t really understand those three problems until you understand in detail why a cookbook formulaic approach would be disastrous incorrect. Remember, the GMAT specializes in producing Quant problems that punish people who try to solve them buy plugging into simple formulas.

Does all this make sense?

Mike 🙂

Sure does, thanks again Mike.

Dear Mayank: You are quite welcome, my friend! 🙂 Best of luck to you!

Mike 🙂