# Combinations Question - Help

• November 23rd 2008, 12:53 PM
cnmath16
Combinations Question - Help
23. From a group of 6 women and 4 men, determine in how many ways a committee of 4 people can be selected with

a) no restriction

C (10,4) = 10! ÷ 6! 4! = 210 ways

b) 4 women

c) 3 women and 1 man

d) 2 women and 2 men

e) 4 men

.. if someone could do b and c for me... I will figure out how to do the rest on my own.
Thanks!
• November 23rd 2008, 12:57 PM
Plato
b) ${6 \choose 4}$

c) ${6 \choose 3}{4 \choose 1}$
• November 23rd 2008, 01:10 PM
cnmath16
Therefore.. the answer to D) will be ( 6 2) ( 4 2)

and the answer to e will be.... (6 0) (4 4) -- for an answer of one? Or is this a special case since all of the men are being used for the selection. The answer of just 1 does not seem correct to me. I was thinking more along the lines of 4! for an answer of 24?