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Suppose a man has 5 aunts and 6 uncles and his wife has 6 aunts and 5 uncles. In how many way's can he call a dinner party of 3 men and 3 woman so that there are exactly 3 of the man's relative and 3 of the wife's ?

I solves this question but I have to consider all the situation's like 3 aunt's and 0 uncles from man's side and 0 aunt and 3 uncles from the wife's side and so on. So through this the process went lengthy so I wanted and simpler solution to this problem if there is any.



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Note that if the man calls $a$ aunts, then he must call $3-a$ uncles. His wife must then call $a$ uncles and $3-a$ aunts. Can you finish it off from here?

Scroll over the gray area for the complete solution.

The number of ways for a fixed $a$ is$$\dbinom{5}a \dbinom{6}{3-a} \dbinom{5}a \dbinom{6}{3-a}$$Now $a$ can take any value from $0$ to $3$. Hence, the total number of ways is$$\sum_{a=0}^3 \dbinom5a^2 \dbinom6{3-a}^2$$

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I will just give it a try.(but not sure or it may be the way in which you did )

There are 11 males and 11 females in total. There are four ways in which 3 men can be selected from the 11 males regarding from man's family or wife's family like 3 from man's and 0 from wife's or 2 from man's and 1 from wife's etc. The no. of ways in which the females can be selected regarding from man's family or wife's family in each of these situations is fixed to satisfy 3 from man's and 3 from wife's. So, finding the combinations for individual selections in the case of selecting womens is enough.

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you have missed to read a point in the ques – Dimensionless Sep 15 '13 at 2:42
can you please tell which part? – Rajath Krishna R Sep 15 '13 at 3:03
I asked for simplar verson this the same one I have used – Dimensionless Sep 15 '13 at 3:07
that's why I have put in the brackets at the beginning that it may be the one that you already know or have used....... – Rajath Krishna R Sep 15 '13 at 3:09
sorry I didn't noticed that one haha – Dimensionless Sep 15 '13 at 3:15

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