# Math Help - Combinatio Help

1. ## Combinatio Help

Suppose that a club consists of 10 men and 13 women. The club is going to form a committee of 8 people. How many 8 person committees have more women than men?

I thought maybe it was (23!/8!*15!)-(10!/8!*2!) but that does not seem like it gives me the correct answer. Any help would be great!

2. Hello, ezwind72!

I don't believe there is a formula for this problem . . .

Suppose that a club consists of 10 men and 13 women.
The club is going to form a committee of 8 people.
How many 8-person committees have more women than men?

$\begin{array}{ccccc}\text{3 men, 5 women} & {10\choose3}{13\choose5} &=& 154,\!440 \\ \\[-3mm]
\text{2 men, 6 women} & {10\choose2}{13\choose6} &=& 77,\!220 \\ \\[-3mm]
\text{1 man, 7 women} & {10\choose1}{13\choose7} &=& 17,\!160 \\ \\[-3mm]
\text{0 men, 8 women} & {10\choose0}{13\choose8} &=& 1,\!287 \\ \\[-3mm] \hline \\[-3mm]
& & \text{Total:} & 250,\!107 \end{array}$

There are $\boxed{{\color{blue}250,\!107}}$ committees with a majority of women.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

That was such an unusual-looking answer, I just had to run a check.

$\begin{array}{cccc}\text{4 men, 4 women} & {10\choose4}{13\choose4} &=& 150,\!150 \\ \\[-3mm]
\text{5 men, 3 women} & {10\choose5}{13\choose3} &=& 72,\!072 \\ \\[-3mm]
\text{6 men, 2 women} & {10\choose6}{13\choose2} &=& 16,\!380 \\ \\[-3mm]
\text{7 men, 1 woman} & {10\choose7}{13\choose1} &=& 1,\!560 \\ \\[-3mm]
\text{8 men, 0 women} & {10\choose8}{13\choose0} &=& 45 \\ \\[-3mm] \hline \\[-3mm]
& & \text{Total: } & 240,\!207 \end{array}$

Hence, there are: . $250,\!107 + 240,\!207 \:=\:{\color{blue}490,\!314}$ possible committees.

Check . There are: . ${23\choose8} \;=\;{\color{blue}490,\!314}$ possible committees . . . YAY!

3. I see what you did there. I was wayyyyyyy off! But I now understand how you came about that answer. The formula using summation notation would be (I wish I could say I came up with that solely on my own).

$\sum\limits_{k = 5}^8 {{13 \choose k}{10 \choose 8-k} }$