For some questions on the GMAT, you will need to know the Combinations formula.

Combinations formula =** n! / k! (n-k)!**

**n = the bigger number (what we’re choosing from) **

**k = the smaller number (how many we’re choosing)**

Check out this question from Math Revolution:

How many committees can be formed comprising 2 male members selected from 4 men, 3 female members selected from 5 women, and 3 junior members selected from 6 juniors?

A. 900

B. 1200

C. 1500

D. 1800

E. 2400

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We could almost rephrase this as three different questions:

How many ways to choose 2 from 4?

How many ways to choose 3 from 5?

How many ways to choose 3 from 6?

Let’s start with the first question: *how many ways to choose 2 from 4? *

4! / 2! (4-2)!

4! / 2!2!

(4 x 3 x 2 x 1) / (2 x 1)(2 x 1)

24/4 = 6

But if we counted this and said the four males were ABCD, we could just list AB, AC, AD, BC, BD, and CD, and you can see it also equals 6.

Anyway, finding the other two:

How many ways to choose 3 from 5?

5! / 3! (5-3)!

5! / 3! 2!

We can cancel out 3! from both numerator and denominator.

5 x 4 / 2

20/2 = 10

How many ways to choose 3 from 6?

6! / 3! (6-3)!

6! / 3! 3!

We can cancel out one 3! from both numerator and denominator.

6 x 5 x 4 / 3 x 2

120 / 6 = 20

Back to our original questions:

How many ways to choose 2 from 4? 6

How many ways to choose 3 from 5? 10

How many ways to choose 3 from 6? 20

Now, multiply all those numbers together!

6 x 10 x 20 = 1200

The correct answer is (B).