# Math Help - Counting, Inclusion-Exclusion Rule

1. ## 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: $7*(nCr(16,4) - nCr(15,3)) = 9555$ (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: $7*9*(nCr(16,3)-2*nCr(15,2)+nCr(14,1)) = 22932$ (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 $7*9*nCr(14,3)$, 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

2. 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?

$\displaystyle\text{With no restrictions, there are: }\:{16\choose5} \:=\:\frac{16!}{5!\,11!} \:=\:4368\text{ possible committees.}$

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

$\text{There are: }\:{9\choose5} \:=\:126\text{ all-male committees.}$

Therefore, there are: . $4368 - 126 \:=\:4242\text{ committees with at least one woman.}$

(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 $126$ all-male committees.

$\displaystyle\text{We find that there are: }\:{7\choose5} \,=\,21\text{ all-female committees.}$

Therefore, there are: . $4368 - 126 - 21 \:=\:4221\text{ committees}$
. . $\text{with at least one woman and at least one man.}$