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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1-D

2-B

3-E

4-D

5-C

are all correct?

Hi Mane! The answers are in a comment below the problem. The answers are:

1. D

2. B

3. B

4. D

5. C

Hi Chris, could you be please reply whether the answers given below are correct.

1. Since the questions asks for probability that the committee will be exclusive to one gender I take this as an “OR” question (either women or men) where addition must be applied.

P(three women) = 5/10 * 4/9 * 3/8 = 1/12

P(three men) = 5/10 * 4/9 * 3/8 = 1/12

P(three women or men) = P(three women) + P(three men) = 1/6 [ D ]

2. A-L letters are in total 12, and the question says a three-letter code is formed with no letter used more than once and it has to be consecutive.

For the first slot: P(having any one letter from A-L) = 12/12 = 1

For the second slot: P(having a consecutive letter) = 1/11 ( the denominator here is 11 since we don’t account for the letter that has been used for the first sot)

For the third slot: P(having a consecutive letter) = 1/10 ( similar to the second slot)

P(any one letter) * P(consecutive letter) * P(consecutive letter) = 1 * 1/11 * 1/10 = 1/110 [ D ]

3. 10! [ E ]

Since words cannot be used more than once, the number of choice each slot has decays by 1.

4. For this question, I assumed x and n and by this way I ended up with answer [ B ]

Lets assume n = 3, and x = 5

Since the question asks for unique team, this would be a combination problem

For column A:

5C3 = 10

For column B:

6C4 = 15

Hence the answer [ B ]

5.Assuming x = 3

12C3 = 220

Only when p equals choices (b,d) the total number of lunar missions remain unchanged while choices a,c,e changes. Anyway, I guess this might be wrong.

Hi Rajkumar,

The answers are:

1. D

2. B

3. B

4. D

5. C

You can check your answers against this Urch Forum. . Poster nkt06 was the first to get them all correct (though in his first post he got #3 wrong). I hope this helps!

1. (D) 1/6

2. (B) 1/66

3. (B) 10!/32

4. R>S if(n>x) and R=S if(n=x)

5. (A) x

Problem 2 is not clearly worded. “Three consecutive letters” could mean three such letters in any order, not just a backwards or forwards sequence. For instance, why would CAB not be a valid code? Those are consecutive letters. I kept getting 1/22 for my answer and couldn’t figure out why until I read.yhe answer.

Hi Sarah,

Consecutive numbers are numbers which follow each other in

order, without gaps, from smallest to largest or from what is considered “first to last.” If we apply this meaning to our alphabet letters, we can arrive at the same answer. While I do not agree that the wording of this question is off, I can certainly understand why this is tricky. Hopefully the way of thinking will stick with you and you can conquer other GRE problems moving forward! 🙂Hi Chris,

Could u let us know what are the methods for solving these problems. That link is not working!

Thanks

Hi Sudeepta,

Here is a great resource about how to solve permutations and combinations: https://magoosh.com/gre/2011/gre-math-combinations-and-permutations/

1)d

2)e

3)b

4)B

5 couldnt solve

what is the answer?give with explanations..

Hi Pedrich,

Chris posted this link in the other comments: http://www.urch.com/forums/gre-math/130701-combinations-permutations-challenge-win-free-subscription.html

Happy studying! 🙂

The link doesn’t work

Hi Shubham,

It does work, but something is really weird with the Urch website for me and I have to scroll about 80% of the way down to see the forum. Before that it looks like a ton of error. If this is what you see, too, just keep scrolling! 🙂

Hi! I know this was an old post, but was there ever a video reply with explanations? I saw the thread on urch, but was hoping there was a video explanation as well. Thanks!

1.D

2.B

3.E

4.D

5.C

Chris,

I’m on the advanced math six month plan for the revised GRE. I’m five months out exactly on Thursday. I should have about four weeks to spare because I started the six month plan about 9 months out anticipating that several work related trips would set me back a touch. I’ve completed many Magoosh practice questions as part of the plan. I have answered about 75% of the Math questions I attempt correctly. Lumped together, counting, exponents, and combinations/permutations related questions account for almost 2/3 (%64) of my incorrect questions. I do not score better than 50% on any of these topics individually.

At first, I had resigned myself to the idea that I may have just reached diminishing returns with regard to improving my Quantitative Reasoning score, but given the disparity between my performance in these topics and the others, I have to believe there is a way to improve these particular areas. Are there any specific sources you can recommend for improving these individual areas? Besides the resources listed in the study plan, The only one I could think to try was the Manhattan 5lb book. Any, help is much appreciated.

-Eric

Hi Eric!

Yes, those topics are very confusing! For counting/permutations/combinations, I’m wondering if it goes back to the basics. For combinations/permutations it’s knowing when to use which one. I’d go through as many questions, whether from Magoosh, the MGRE big book, etc. and tell which formula the question calls for (but don’t actually use the math). If you are consistently guessing incorrectly, then it is clearly an issue of conceptualization–not math.

In that case, revisit the lesson videos on combinations/permutations, the MISSISSIPPI rule etc. Once you’ve brushed up on these, then go through the 5 lbs. book doing practice questions. Other sources, include the official GMAT guide. Also, I’ve written plenty of permutations/comb./counting questions on this blog (some of which are quite tricky :)).

Let me know if that helps boost up your percentage. It would be good to see you get near 95% accuracy on all question types :).

1) E

2) B

3) B

4) D

5) C

Hi Billy, to see the answers (and explanations) take a look at the following link posted in one of my comments below.

where is the explanation? I have checked every link

Hi, Dan

Here it is: http://www.urch.com/forums/gre-math/130701-combinations-permutations-challenge-win-free-subscription.html

I hope that helps! Let us know if you have any other questions! 🙂

Best,

Margarette

1.D

2.D

3.B

4.B

5.C

please provide solutions for the examples

Here is a useful thread on Urch that has several different approaches to each of the problems above (including the right answers.) Make sure you scroll to the second page to see all of the comments.

http://www.urch.com/forums/gre-math/130701-combinations-permutations-challenge-win-free-subscription.html

i got first four answers

1. 1/6

2. 1/66

3. 10!/32

4. A is greater than B

you are the only right person i have encountered so far who has given a right answer for the question number 3….

count me in lmao. it’s all about practice

1 D

2 1/22 is what I got,none of the choices give that

3.No clue

4. Column B is greater

5.C

1:D

2;B

3;E

4:B

5:C

i got no. 1 (D)…I will keep working on it. thanks for the post!