I know a family that has $10$ children and half of them are married.
Sometimes they ask me about mathematics, and ask me to show them nice things...
I asked them: How many possible ways there are to pick $5$ of your children and that $3$ of them (at our selection) will be married?
I explain them that this is a problem at combinatorics.

My answer (and I'll be glad if you will correct me): $$\binom{5}{3}\cdot \binom{5}{2}$$ Why?
First of all we choosing $3$ married children from the $5$ that we have - and this is $\binom{5}{3}$.
Than, we have another more $5$ children to choose $2$ from them, and this is why we have $\binom{5}{2}$.

If the question was that we can have more then $3$ married children so the answer will be:$$\binom{5}{3}\cdot \binom{7}{2}$$ Why?
Because we move the unchosen married children to the group of the unmarried children...

At the next section I asked the same thing but I added one more thing - that the minimum of married children can be $1$ and the maximum is $4$ (e.g. at $2$ there can only be $2$ married children and no more... IF there can be more it will $7$ at the right binom and 5), my answer is: $$\sum_{i=1}^{4}\binom{5}{i}\cdot \binom{5}{5-i}$$ I'm right?

I just want to tell you that I didn't show my answers to the family, I put here to know if I'm right...

Thank you!


The first and the third are right, but the second is not right.

For the second, let $A,B,C,D,E$ are married children and let $F,G,H,I,J$ are unmarried children.

Then, $\binom{5}{3}\cdot \binom{7}{2}(=210)$ counts a set $\{ABCDE\}$ multiple times.

1) First you choose $A,B,C$ and you choose $D,E$.

2) First you choose $A,C,D$ and you choose $B,E$.

and so on...

So, the answer is $$\binom{5}{3}\binom{5}{2}+\binom{5}{4}\binom{5}{1}+\binom{5}{5}\binom{5}{0}=100+25+1=126$$ in the same way as the first and the third.

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  • $\begingroup$ Thank you! You are talking about this? en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle (At the second one). $\endgroup$ – CS1 Jan 18 '14 at 18:12
  • $\begingroup$ And why I count $\left\{ABCDE\right\}$ multiple times? Because the second the to married children moves to the other group.... Thank you! $\endgroup$ – CS1 Jan 18 '14 at 18:14
  • $\begingroup$ Well, not really. Also, I've already explained it. Do you see the points 1) 2) I wrote? $\endgroup$ – mathlove Jan 18 '14 at 18:16
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    $\begingroup$ You mean the number of married children is either $3,4,$ or $5$, right? So, consider each case separately. $\endgroup$ – mathlove Jan 18 '14 at 18:41
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    $\begingroup$ Yes. $\binom{5}{3}\binom{5}{2}$ represents the case in which the number of married children is $3$. $\binom{5}{4}\binom{5}{1}$ represents the case in which the number of married children is $4$. $\binom{5}{5}\binom{5}{0}$ represents the case in which the number of married children is $5$. $\endgroup$ – mathlove Jan 18 '14 at 18:43

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