Hi there,

So here's a problem I've got for homework...

So for part A I have..Seven women and nine men are on the faculty in the mathematics department at a school.

a) How many ways are there to select a committee of 5 members of the department if at least one woman must be on the committee?

b) How many ways are there to select a committee of 5 members if at least one woman and one man must be on the committee?

So the calculation is: $\displaystyle 7*(nCr(16,4) - nCr(15,3)) = 9555$ (final answer)--> 7 ways to fill first slot with a woman we will call w.

--> nCr(16,4) ways to fill the next 4 slots

-----> Minus the combinations containing w = nCr(15,3)

And for part B...

So the calculation is: $\displaystyle 7*9*(nCr(16,3)-2*nCr(15,2)+nCr(14,1)) = 22932$ (final answer)--> 7 ways to fill first slot with a woman we'll call w

--> 9 ways to fill second slot with a man we'll call m

--> nCr(16,3) ways to fill the next 3 slots

-----> Minus combinations containing w = nCr(15,2)

-----> Minus combinations containing m = nCr(15,2)

-----> Plus combinations containing w and m = nCr(14,1)

These answers just don't seem reasonable... especially since it looks like the additional restrictions that are applied in part B actually result in MORE possible ways to choose the committee instead of less...

I also tried considering that, in part B, instead of choosing 3 people from 16 to fill the remaining slots, you simply subtract 2 from the number of people (because you exclude w and m from the set of people available to fill the slots), thus the calculation becomes simply $\displaystyle 7*9*nCr(14,3)$, but this yields the exact same answer of 22932. The same applies to part A when I tried applying this logic.

Where am I going wrong?

Thanks