Permutations
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
Combinations
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 threeperson 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.
Practice Questions
1) A bookseller has two display windows. She plans to display 4 new fiction books in the left window, and 3 new nonfiction books in the right window. Assuming she can put the four fiction books in any order, and separately, the three nonfiction books in any order, how many total configurations will there be for the two display windows?

(A) 24
(B) 72
(C) 144
(D) 336
(E) 420
2) The countymandated 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?

(A) 15
(B) 30
(C) 90
(D) 150
(E) 360
Answers and Explanations
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 nonfiction 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.
Thank you Magoosh.
The second question can also be solved in a simple way
# of combinations of novels + # of combinations of plays
= 3!/1*2! + 5!/2!*3!
= 3 * 10
= 30
Neeraj
Equation should be
= 3!/1*2! * 5!/2!*3!
It is multiplication of the two combinations not the addition of the two.
Very helpful posts. However, I have a doubt. Although I kind of understand why in the second question, you went with 3C1 * 5C2 = (3*10)= 30, I am confused about one thing.
Another way to do solve this problem would be to make three boxes– first box for the novel, second for a play, third also for a play.
The first box can be filled in 3 ways; the second in 5 ways, the third in 4 ways.
So total no. of combinations: 3*5*4 = 3*20=60. It gives us a different answer. Why is this wrong? Am I missing something about “order matters, order does not matter”.
I suppose in this case, order does not matter. But I don’t know how to translate that into the Math here. Furthermore, I consulted your earlier post on P&C (http://magoosh.com/gmat/2012/gmatquanthowtocount/). In the Shakespeare example, where you say:
“For example: Shakespeare wrote fifteen comedies, ten histories, and twelve tragedies. If we are going to pick one of each kind, and ask how many different trios of plays can we create, the total number is simply 15*10*12. ”
How is the Shakespeare problem above different from this question here? I would be most grateful for your explanation.
Dear Siddharta,
I’m happy to respond. This is a very tricky think about counting problems. Your method was perfectly valid, but you overlooked one thing: repetitions.
Let’s call those five Shakespeare plays {A, B, C, D, E}. You said: five choices for the first play, and four choices for the second play. That’s perfectly true, but each pair gets counted twice. If we pick A first, and B second, or if we pick B first, and A second, either results in the same pair (A, B) on the reading list. Order doesn’t matter here, but in your calculation, you presumed that order mattered, so you wound up doublecounting, which gave you an answer twice as big as the correct answer. All of these issues are explained in detail in the Magoosh lessons on counting.
Does all this make sense?
Mike
Hi Mike,
I’ve come across your blog 24 hours before my gmat. and it feels godsent. Just reading through these simple explanations with bits of humor thrown in is calming me a lot
I had one simple question I get confused about differentiating between permutation and combination sometimes. Is there a blog post explaining that?
THanks so much
Hi Ally,
We’re glad to hear you feel that way!
Check out this blog post about permutations: http://magoosh.com/gmat/2012/gmatpermutationsandcombinations/
Best of luck on your GMAT!
Jessica
I’m so grateful to this site. Not that i have secured a score yet, but the prep time is so much fun. thanks to magoosh.
such little details explained so explicitly. sometimes I couldnt even fathom where i was going wrong. thanks to you, i realized depending too much on formulas too can be an error.
thankyou again! keep up!
cheers
what are example of companies that uses permutation or combination to perform their inventory management or other business function
Dear Kimberly,
I’m happy to respond. First of all, understand that you are asking the wrong person. By trade, I am a GMAT expert: I can tell you cartloads about how this math appears on the GMAT. I am NOT an industry specialist in any industry, so I am certainly not equipped to discuss details of how this math might be used in daytoday operations of specific businesses. I have my guesses, but I have no data at my fingertips on that.
Here’s what I will say. I could easily imagine the following scenario playing out many times. The CEO or CFO or other high muckamuck at a corporation wants to know how many ways they can market something or package something, or some similar question — what we would recognize as a combinations question. The high muckamuck doesn’t know this, so that person sends the question to some underling, who asks someone else, and eventually the question finds it way to someone such as an engineer who realizes that it’s actually a very easy question to answer when one knows the math. This engineer produces the correct answer, and this answer migrates back up the food chain until it reaches the high muckamuck who asked in the first place. Now, here’s the funny thing. If we surveyed the company, that is, if we asked the high muckamuck or one of the immediate underlings, “Do you use permutations and combinations?”, their answer might well be: “What are those? I’ve never even heard of them!” Even though someone at the company actually did use permutations and combination to answer the question, the folks who received the answer may not be aware of that in the least. People who run companies and their press agents often have lots of numbers at their disposal, and they don’t always understand exactly how every last number was calculated: after all, knowing that calculation is not part and parcel of their job. All this would make it extremely problematic to determine which companies are actually using this math.
All this is to say that is you have all the math at your fingertips, you will walk through the corporate works with a kind of magic power, able to compute things quickly that baffle others.
Does all this make sense?
Mike
Greetings Mike,
I wanted to thank you for the great work and explanations you have provided us. I have been studying for the GMAT for a while, and there are some concepts in my prep book that don’t thoroughly explain on how to get certain types of questions. The addition of your material has def helped in a significant way.
Kind Regards,
Herpal Pabla
Dear Herpal,
My friend, you are more than welcome. I am very glad to hear that you found this helpful! Best of luck to you in the future!
Mike
hi,
Could you please provide some more practice questions based on the above concept. Thanks!
Dear Shilpi
Take a look at this post:
http://magoosh.com/gmat/2013/difficultgmatcountingproblems/
Mike
Thanks a lot
Dear Shilpi,
You are quite welcome. Best of luck to you!
Mike
“With the plays, we are taken a combination of 3 from a set of 5”
Should be 2 from a set of 5. Otherwise, a very good post.
Steven,
Thanks for catching that — I just fixed it. Incidentally, I probably made the mistake because, as you may know, 5C2 = 5C3 — at a certain level of analysis, there’s no difference picking two from five or three from five. But for clarity, it’s good to make this change. Thanks,
Mike
Hi Mike
Can we do it this way?
1/3 * 2/5 * 1/4 = 1/30
so we have 30 combination?
Thanks
Naren,
I assume you mean practice question #2. Yes, that would be another way to compute the number of possible reading lists.
Mike
Mike – Thanks for the valuable information. I have a clarification to make with respect to question 1.
I was thinking – would not the answer be 24 + 6 = 30. I am thiking as she wants the ‘ficiton’ books still too be placed on the left hand side in any order and the same goes for ‘non fiction books.
Am i missing a trick here ?
Rahul Sehgal
GMAT aspirant !!
Dear Rahul,
The big conceptual piece you are missing is the Fundamental Counting Principle. See this blogpost:
http://magoosh.com/gmat/2012/gmatquanthowtocount/
Let me know if that doesn’t clear up why we *multiply* instead of adding.
Mike
A nice intro to P&C and, probably you are true WRT the GMAT, this is probably more than enough to crack
Thank you very much.
Mike