# Math Help - Clarifying the first combination example of this text.

1. ## Clarifying the first combination example of this text.

From a group of 5 women and 7 men, how many different committees consisting of 2 women and 3 men can be formed? What if 2 of the men are feuding and refuse to serve on the committee together?

For the second question the book says that there are 5 total invalid combinations, $\binom{2}{2}\binom{5}{1}=5$, but then provides no explanation as to how it arrived at that conclusion.

Could someone explain how this poorly executed textbook arrived at this conclusion for me?

2. ## Re: Clarifying the first combination example of this text.

Originally Posted by bkbowser
From a group of 5 women and 7 men, how many different committees consisting of 2 women and 3 men can be formed? What if 2 of the men are feuding and refuse to serve on the committee together?
For the second question the book says that there are 5 total invalid combinations, $\binom{2}{2}\binom{5}{1}=5$, but then provides no explanation as to how it arrived at that conclusion.
With no restrictions at all there are $\binom{5}{2}\binom{7}{3}$ ways to form the committee.

Put two feuding men altogether there are $\binom{5}{2}\binom{5}{1}$ ways to form the committee.

Now subtract the second from the first.

3. ## Re: Clarifying the first combination example of this text.

Originally Posted by Plato
With no restrictions at all there are $\binom{5}{2}\binom{7}{3}$ ways to form the committee.

Leaving the two feuding men off altogether there are $\binom{5}{2}\binom{5}{3}$ ways to form the committee.

Now subtract the second from the first.
So $\binom{2}{2}\binom{5}{1}$ is a simplification of $\binom{5}{2}\binom{5}{3}$?

I'm getting;

$\binom{5}{2}\binom{5}{3}=\frac{5!}{(5-2)!2!}\frac{5!}{(5-3)!3!}=2\frac{5!}{3!2!}=20\neq5$

?

4. ## Re: Clarifying the first combination example of this text.

Op sorry there's an edit.

5. ## Re: Clarifying the first combination example of this text.

Still getting something wierd.
$\binom{5}{2}\binom{5}{1}=\frac{5!}{(5-2)!2!}\frac{5!}{(5-1)!1!}=\frac{5!5!}{3!2!4!}=100\neq5$

6. ## Re: Clarifying the first combination example of this text.

I think you may be misunderstanding the book.
An "invalid" committee is one in which those two specific men serve together. Treat those two men as a single unit. There is one way, $\begin{pmatrix}2 \\ 2 \end{pmatrix}$ to choose that unit and $\begin{pmatrix}5 \\ 1\end{pmatrix}$ ways to choose the third man. So there are $\begin{pmatrix}2 \\ 2 \end{pmatrix}\begin{pmatrix}5 \\ 1\end{pmatrix}$ to choose the men for an "invalid" committee. But there will still be $\begin{pmatrix}5 \\ 2 \end{pmatrix}$ ways to choose the women and so $\begin{pmatrix}5 \\ 2\end{pmatrix}\begin{pmatrix}2 \\ 2 \end{pmatrix}\begin{pmatrix}5 \\ 1\end{pmatrix}$ "invalid" committees.

7. ## Re: Clarifying the first combination example of this text.

Originally Posted by bkbowser
Still getting something wierd.
$\binom{5}{2}\binom{5}{1}=\frac{5!}{(5-2)!2!}\frac{5!}{(5-1)!1!}=\frac{5!5!}{3!2!4!}=100\neq5$
Sorry that should be 50 not 100

8. ## Re: Clarifying the first combination example of this text.

Originally Posted by HallsofIvy
I think you may be misunderstanding the book.
An "invalid" committee is one in which those two specific men serve together. Treat those two men as a single unit. There is one way, $\begin{pmatrix}2 \\ 2 \end{pmatrix}$ to choose that unit and $\begin{pmatrix}5 \\ 1\end{pmatrix}$ ways to choose the third man. So there are $\begin{pmatrix}2 \\ 2 \end{pmatrix}\begin{pmatrix}5 \\ 1\end{pmatrix}$ to choose the men for an "invalid" committee. But there will still be $\begin{pmatrix}5 \\ 2 \end{pmatrix}$ ways to choose the women and so $\begin{pmatrix}5 \\ 2\end{pmatrix}\begin{pmatrix}2 \\ 2 \end{pmatrix}\begin{pmatrix}5 \\ 1\end{pmatrix}$ "invalid" committees.
OK, that does make sense to me.

Thanks.