Question

Check my work please?

If I have 8 men and 9 women, and Im making a guest list. How many ways can I invite a list of 5 men and 5 women, given that: If one of the men, Carl, goes, one of the women, Alice, will refuse to go, and vice versa?

Answer 1

My thoughts are that, I have 2 sub-cases to consider, and the number of unique lists will be their sum:
(all the lists in which Carl goes, and Alice doesn't) + (all the lists in which Alice goes and Carl doesn't).

For the first part. If I have 8 total men, but this is all the cases that Carl goes, so, I believe I have (7 choose 4) for the men and and (8 choose 5) for the women.

Answer 2

for the 2nd sub-case, it would be (7 choose 5) for the men and (8 choose 4) for the women.

Answer 3

\[\left(\begin{matrix}7 \\ 4\end{matrix}\right) * \left(\begin{matrix}8 \\ 5\end{matrix}\right) + \left(\begin{matrix}7 \\ 5\end{matrix}\right) * \left(\begin{matrix}8 \\ 4\end{matrix}\right)\] = 3430...

Answer 4

I feel like I am missing something here...

Answer 5

Well the total amount of combinations is 8C5 * 9C5.

Now you need to discount the combination when 1 of the men and 1 of the women doesn't get chosen.

Answer 6

So something like, (# of lists with bob) - (# of lists without bob) times (# lists with alice) - (# of lists without)

Answer 7

which I guess is:\[[\left(\begin{matrix}7 \\ 4\end{matrix}\right) - \left(\begin{matrix}7 \\ 5\end{matrix}\right) ]*[ \left(\begin{matrix}9 \\ 4\end{matrix}\right) - \left(\begin{matrix}9 \\ 5\end{matrix}\right)]\]

Answer 8

I'm pretty sure this is the answer.
You choose Carl that is 8C1 = 8 ways to choose Carl!. and multiply this by 9C5 (all the women combinations)

Do the same with Alice
Choose alice 9C1 = 9 ways to choose alice from 9 women. Multiply this by 8C5 men.

8C5 * 9C5 - 8*9C5 - 9*8C5

Answer 9

ahhh ok. I think I understand that.