Math Help - Counting, Inclusion-Exclusion Rule

Hi there,

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

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 for part A I have..

--> 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)

So the calculation is: (final answer)

And for part B...

--> 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)

So the calculation is: (final answer)

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 , 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

Hello, Fox2013!

Seven women and nine men are on the faculty.
How many ways are there to select a committee of five people

(a) if at least one woman must be on the committee?

The opposite of "at least one woman" is "no women" (all men).



Therefore, there are: .

(b) if at least one woman and one man must be on the committee?

We must not have all-male committees nor all-female committees.

In part (a), we found that are all-male committees.



Therefore, there are: .
. .