Complicated combinatorics problem

[Zermelo]

Hello,
I'm working on a complicated combinatorics problem from my Discrete Math class. The problems goes like this:
The husband has 12 cousins: 5 men and 7 women
His wife also has 12 cousins: 7 men and 5 women.
They don't have any common cousins.
They want to invite 6 husband's cousins and 6 wife's cousins, but such that there will be 6 men and 6 women at the party (excluding the husband and wife).
How many ways are there to do that?

Well, I drew myself a sketch and managed to "brute force it", what I got is:

[MATH]{7\choose 6}{7\choose 6} + {7\choose 5}{5\choose 1}{5\choose 1}{7\choose 5} + {7\choose 4}{5\choose 2}{5\choose 2}{7\choose 4} + {7\choose 3}{5\choose 3}{5\choose 3}{7\choose 3}+ {7\choose 2}{5\choose 4}{5\choose 4}{7\choose 2} + {7\choose 1}{5\choose 5}{5\choose 5}{7\choose 1}[/MATH]
My logic was: the first solution to invite all the women from the husband's family and all the men from the wife's, the second solution is to invite 6 women from the husband's and 1 from the wife's and that mean's 1 man from the husband's and 5 from the wife's etc etc.
My first question is: is my line of reasoning ok? Is this a correct use of the product rule?
The second question is: is there a simpler way of doing this? Maybe I can translate this problem to combinations or permutations with repetition? Or is this the only way?
Thanks in advance!

 

[pka]

Hello,
I'm working on a complicated combinatorics problem from my Discrete Math class. The problems goes like this:
The husband has 12 cousins: 5 men and 7 women
His wife also has 12 cousins: 7 men and 5 women.
They don't have any common cousins.
They want to invite 6 husband's cousins and 6 wife's cousins, but such that there will be 6 men and 6 women at the party (excluding the husband and wife).
[MATH]{7\choose 6}{7\choose 6} + {7\choose 5}{5\choose 1}{5\choose 1}{7\choose 5} + {7\choose 4}{5\choose 2}{5\choose 2}{7\choose 4} + {7\choose 3}{5\choose 3}{5\choose 3}{7\choose 3}+ {7\choose 2}{5\choose 4}{5\choose 4}{7\choose 2} + {7\choose 1}{5\choose 5}{5\choose 5}{7\choose 1}[/MATH]

\(\sum\limits_{k = 0}^5\dbinom{7}{6-k}\dbinom{5}{k}\dbinom{5}{k}\dbinom{7}{6-k}=267148\) See HERE

 

[JeffM]

Hello,
I'm working on a complicated combinatorics problem from my Discrete Math class. The problems goes like this:
The husband has 12 cousins: 5 men and 7 women
His wife also has 12 cousins: 7 men and 5 women.
They don't have any common cousins.
They want to invite 6 husband's cousins and 6 wife's cousins, but such that there will be 6 men and 6 women at the party (excluding the husband and wife).
How many ways are there to do that?

Well, I drew myself a sketch and managed to "brute force it", what I got is:

[MATH]{7\choose 6}{7\choose 6} + {7\choose 5}{5\choose 1}{5\choose 1}{7\choose 5} + {7\choose 4}{5\choose 2}{5\choose 2}{7\choose 4} + {7\choose 3}{5\choose 3}{5\choose 3}{7\choose 3}+ {7\choose 2}{5\choose 4}{5\choose 4}{7\choose 2} + {7\choose 1}{5\choose 5}{5\choose 5}{7\choose 1}[/MATH]
My logic was: the first solution to invite all the women from the husband's family and all the men from the wife's, the second solution is to invite 6 women from the husband's and 1 from the wife's and that mean's 1 man from the husband's and 5 from the wife's etc etc.
My first question is: is my line of reasoning ok? Is this a correct use of the product rule?
The second question is: is there a simpler way of doing this? Maybe I can translate this problem to combinations or permutations with repetition? Or is this the only way?
Thanks in advance!

Email the cousins pictures of the opposite cousins of opposite sex and ask each who is attractive. It will be an interesting party. There was a French movie about it: Cousin, Cousine