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There are 10 players on the basketball team, 12 players on the volleyball team, and 15 members of the track team. Of those players, 2 athletes are on all three teams. How many committees of 4 players can be formed if there must be at least 1 member from each team represented? Order doesn't matter.

attempt:

I got 33 because I subtracted 2 from each team, and then added it to the total (31+2 = 33)

33C4 - 23C4 - 21C4 - 18C4 - 8C4(all bball) - 10C4(all vball) - 13C4(all tennis)

= 22,025

But apparently the answer is 24,015

Am i doing something wrong?

Jessica
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1 Answers1

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There are $2$ versatiles. We will assume that apart from that, there is no overlap between teams.

We interpret the problem as meaning that if one or two of the versatiles is on the committee, then the condition that every team is represented is fulfilled.

There are $31$ non-versatiles.

Committees with $2$ versatiles: $\binom{31}{2}$.

Committees with $1$ versatile: $(2)\binom{31}{3}$.

Finally, we must count the committees with no versatiles. There are respectively $8$, $10$, and $13$ on the various teams. With no restrictions that gives $\binom{31}{4}$ committees. From this we must subtract the number of forbidden committees. The number of forbidden committees is $$\binom{18}{4}+\binom{23}{4}+\binom{21}{4}-\binom{8}{4}-\binom{10}{4}-\binom{13}{4}.$$ (Be careful about the minus signs!)

Remark: We computed using the above expressions (well, the real work was done by a battery powered assistant). The numerical result is $24015$. So the interpretation mentioned at the beginning of the answer was probably the intended one.

André Nicolas
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  • but in the 31C4, you are missing the 2 extra people – Jessica Feb 24 '14 at 07:04
  • that are on all three teams – Jessica Feb 24 '14 at 07:04
  • and why are you only adding the beginning part and subtracting the end? arn't you suppose to add up all the forbidden committees and then subtract them from the total? – Jessica Feb 24 '14 at 07:05
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    The number of allowed committees with non-versatiles only is $\binom{31}{4}-\binom{18}{4}-\binom{23}{4}-\binom{21}{4}+\binom{8}{4}+\binom{10}{4}+\binom{13}{4}$. You have Inclusion/Exclusion wrong. When we are counting the forbiddens, and calculate $\binom{18}{4}+\binom{23}{4}+\binom{21}{4}$, we are double-counting the single sport committees. So for the forbidden committees, we must subtract the single sport ones. Informally, as a menemonic, in Inclusion/Exclusion, signs will alternate. – André Nicolas Feb 24 '14 at 07:17
  • Note the typo in the second line, it is supposed to be $\binom{10}{4}$, but I can't find the TeX error in the limited time allowed for editing. – André Nicolas Feb 24 '14 at 07:22
  • Reply to your first comment: The committees that involve one or more of the two extra people have been counted separately, because they are "special" as far as fulfilling the condition goes. The count for these given in the answer is $\binom{31}{2}+2\binom{31}{3}$. – André Nicolas Feb 24 '14 at 07:34
  • Thank you for the very thorough explanation! – Jessica Feb 25 '14 at 01:38