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?
For this question, it's easier if you know the Combinations formula, but you could technically count.
Combinations formula = n! / k! (n-k)!
n = the bigger number (what we're choosing from)
k = the smaller number (how many we're choosing)
For example: 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
_________________
Vivian Kerr | GMAT/GRE Tutor @ http://www.gmatrockstar.com | gmatrockstar[at]gmail.com | solid 5-star reviews on Yelp!
Former Kaplan and Grockit instructor. I've freelanced with Veritas Prep, Magoosh, and most of the bigger test prep companies. Now offering Skype-based private tutoring for the GMAT and GRE.
Reading Comprehension is my jam!