# Supposed that a department contains 10 men and 15 women. How many ways are there to form a committee with six members if it must have more women than men?

##### 1 Answer

There are

#### Explanation:

First we ask, how many ways are there to choose

The answer turns out to be

#((n),(r)) = frac{n!}{r!(n-r)!}#

So for example, how many ways are there to choose

The answer is

#((10),(2)) = frac{10!}{2! * 8!} = 45#

Now, with the condition that there must be more women than men, we are left with only 3 options: 4w+2m, 5w+1m and 6w+0m.

Number of ways with 4 women and 2 men

#((15),(4)) * ((10),(2)) = frac{15!}{4! * 11!} * frac{10!}{2! * 8!} = 1365xx45#

# = 61425#

Number of ways with 5 women and 1 man

#((15),(5)) * ((10),(1)) = frac{15!}{5! * 10!} * frac{10!}{1! * 9!} = 3003xx10#

# = 30030#

Number of ways with 6 women and 0 man

#((15),(6)) * ((10),(0)) = frac{15!}{6! * 9!} * frac{10!}{0! * 10!} = 5005xx1#

# = 5005#

Now all you have to do is to add the 3 cases up.

# 61425 + 30030 + 5005 = 96460#