counting help

zpwnchen · #1 $Old$ November 3rd, 2009, 09:02 PM

Quote:

From a group of 7 women and 6 men a committee consisting of 3 men and 4 women is to be formed. How many different committees are possible if
a) 2 of the women refuse to serve together?
b) 1 man and 1 woman refuse to serve together?

help me

Grandad · #2 $Old$ November 4th, 2009, 03:47 AM

Hello zpwnchen

Quote:

Originally Posted by zpwnchen $View Post$

help me

With no restrictions there are $\binom74=35$ ways of choosing the women and $\binom63=20$ ways of choosing the men; i.e. $35\times20=700$ ways altogether.

a) We now find the number of ways of forming the committee to include the two problem women. With these women already selected we need $2$ more women from the remaining $5$ and $3$ men from $6$ . That's $\binom52\times\binom63=200$ ways. So, excluding these choices, there remain $500$ possible ways.

b) Do this in the same way. Once the problem people have been selected, work out the number of possible choices from the remaining two groups, and subtract from $700$ .

Grandad