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Let's say that there is a team of 28 girls, out of which 8 are 10 years old, 11 are 11 and 9 are 12 years old. In how many ways can we create a team of 6 such that players of every age are represented?

The text book solution is: $299046$ ways

However my solution is $1821600$ ways

My Work:

We know that there are 3 known players in each of possible teams (one from each age group). There are $8*11*9$ ways to choose those 3, and then there are $\binom{25}{3}$ ways to choose the rest.

Am I wrong?

Jakov Gl.
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  • You have counted ways to select three leaders from each age group, and three other players (of any age). That is not what you want. – Graham Kemp Nov 01 '22 at 00:28
  • Well, what if the first three players are $ABC$, the last three are $DEF$. However, if $DEF$ are all different ages, then wouldn't $DEFABC$ also be included? – bobeyt6 Nov 01 '22 at 00:29
  • Likewise Suppose $DEF$ are all ten-years, like $A$. Why was $A$ counted specifically? – Graham Kemp Nov 01 '22 at 00:32
  • Yeah ok that makes sense. So how would you solve it then? – Jakov Gl. Nov 01 '22 at 00:34
  • Since numbers here are so small, doing it by cases is fine. If the numbers were much larger, you should do it by inclusion-exclusion over the events that one of the age groups was missing. – JMoravitz Nov 01 '22 at 00:35
  • I'd recommend using PIE for this one. There are fewer binomials to evaluate that way. – Graham Kemp Nov 01 '22 at 00:46

2 Answers2

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To elaborate on my suggestion of inclusion-exclusion... let $A$ be the set of arrangements with no 10yearolds, $B$ the set with no 11yearolds, and $C$ with no 12yearolds.

You want $|A^c\cap B^c\cap C^c|$, that is the number of arrangements with at least one 10yearold and at least one 11 year old, etc.

You can see this is equal to $|(A\cup B\cup C)^c|$ which expands as $|\Omega|-|A\cup B\cup C|=|\Omega|-|A|-|B|-|C|+|A\cap B|+|A\cap C|+|B\cap C|-|A\cap B\cap C|$

$$\binom{28}{6}-\binom{20}{6}-\binom{17}{6}-\binom{19}{6}+\binom{9}{6}+\binom{11}{6}+\binom{8}{6}-0$$

Completing the arithmetic gives the final answer.

JMoravitz
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  • This gives the same result as in my textbook. – Jakov Gl. Nov 01 '22 at 00:53
  • @JMoravitz, Respected Mr. Moravitz, what does $-0$ mean at the end of the expression $\binom{28}{6}-\binom{20}{6}-\binom{17}{6}-\binom{19}{6}+\binom{9}{6}+\binom{11}{6}+\binom{8}{6}-0$? What the sense does $-0$ have? Explain, please. – Vadim Chernetsov Nov 01 '22 at 15:26
  • @VadimChernetsov In the inclusion-exclusion expansion, we have the $-|A\cap B\cap C|$ term being subtracted at the end. I included $-0$ in the numerical version to emphasize the fact that I had not forgotten this term and that in other problems it is possible for this term to be nonzero and should not be forgotten then. It would be incorrect to assume that the only terms that exist in the expansion are those that are just the singletons $|A|,|B|,|C|$ and the pairs $|A\cap B|,|A\cap C|,|B\cap C|$. The triples, and quadruples, etc... could all exist and be nonzero in future problems – JMoravitz Nov 01 '22 at 15:30
  • It so happens that in this specific problem that term happened to equal zero. I could have not written it if I wanted to make it shorter. I chose to include it intentionally in order to make it clearer how we went from the one line to the next. I could have written it as $\binom{0}{6}$ as well, but I find that to be a greater chance for confusion than what I had done. – JMoravitz Nov 01 '22 at 15:32
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The mistake is that the "all ages" people and the other people are not distinguished. The easiest way I know of is casework.

mathlander
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