Statistics

Question

In a club there are 10 women and 7 men. A committee of 6 women and 4 men is to be chosen. How many different possibilities are there?

Answer 1

 

10 women and 7 men

you want 6 women and 4 men

 

10 women choose 6

n!/r!(n-r)!

10!/(6!)(4!)

(10*9*8*7)/(4*3*2*1)

(10*9*7)/3

(10*3*7)

210 ways to choose the women

 

7 men choose 4

7!/(4!)(3!)

(7*6*5)/(3*2*1)

7*5

35 ways to choose the men

 

210*35=7350 different possibilities

 

It's a combination problem not a permutation problem. For example:

If there are 3 women A, B, and C, how many ways can you choose 2 of them ?

AB, AC, and BC; if A and B are together that is what matters, not who gets picked first or second

A with B is the same as B with A

Answer 2

The other answer is correct for this problem.