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GMAT Permutations and CombinationsBY MIKE MCGARRY  ON JANUARY 11, 2012 IN FORMULAS, PERMUTATIONS AND COMBINATIONS, QUANTITATIVE

PermutationsA permutation is a possible order in which to put a set of objects. Suppose I had a shelf of 5 differ ent books, and I wanted to

know: in how many diffe rent orders can I put these 5 books? Another way to say that is: 5 books have how many diff erent

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

 

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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 differe nt possible three-person 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 el ements

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:

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# 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 non-fiction

books in the right window. Assuming she can put the four fictio n books in any order, and separately, the three non-fiction 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 county-mandated 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 li st 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

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permutations = 4! = (4)(3)(2)(1) = 24

 

The right window will have permutations of the 3 non-fiction 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.

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 About Mike McGarry

Mike creates expert lessons and practice questions to guide GMAT students to success. He has a BS in Physics and an MA in

Religion, both from Harvard, and over 20 years of teaching experience specializing in math, science, and standardized

exams. Mike likes smashing foosballs into orbit, and despite having no obvious cranial deficiency, he insists on rooting for

the NY Mets.

23 Responses to GMAT Permutations and Combinations

Neeraj September 20, 2015 at 10:27 am #

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

REPLY  

Alex September 29, 2015 at 8:52 pm #

Neeraj

REPLY