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The coach of a university soccer team is trying to form a team. From the english department she has found 15 people, and from the cs department she has found 13. She wants to pick the best 14 players for her team

  1. The Coach wants to choose 14 players from her team such that the team includes at least 3 people from the cs and english department. How many ways can she choose the 14 players?

What I have done: What I think I have to do is multiply 15C3, 13C3, and 22C8 (because 6 players were already chosen). Is this the right method?

JMoravitz
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wildpanda
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    No, because that incorrectly applies some sort of significance to whether a person was chosen among the initial three from a department or if they were instead part of the "leftovers." It shouldn't matter to us if a person was picked first or picked last. All that matters is if they are on the team or not. – JMoravitz Oct 20 '22 at 15:48
  • So would I add the combinations instead? – wildpanda Oct 20 '22 at 15:51
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    For a corrected approach, consider breaking into cases based on the total number of people from the English department actually used in the team. Alternatively (since that is a lot of cases) you might instead approach indirectly by counting how many teams would be "bad" because they had two or fewer people from the English department or CS department. – JMoravitz Oct 20 '22 at 15:52
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    If you need more convincing of why your initial approach was wrong, consider a far simpler question of where there was only the English department and no CS department... there were only $14$ people in the English department, and we were looking to make a team of $14$ people from that such that at least three members of the English department were on the team. Stopping and looking at the problem, you should be able to tell immediately there is only one possibility... the team is the entirety of the English department. According to your broken logic, you might have said it was far more. – JMoravitz Oct 20 '22 at 15:56
  • If Im getting what you're saying, there would be 15C14 + 15C13 + 15C12 + 13C13 + 13C12 bad teams correct? – wildpanda Oct 20 '22 at 16:21
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    Almost. $\binom{15}{14}\binom{13}{0}+\binom{15}{13}\binom{13}{1}+\binom{15}{12}\binom{13}{2}+\binom{15}{2}\binom{13}{12}+\binom{15}{1}\binom{13}{13}$. You only chose from the dominant department and did not many any choices about who (if any) came from the other department. – JMoravitz Oct 20 '22 at 16:32
  • Ahh that makes sense! And then to get the answer for possible number of ways you would subtract the value of that expression from 28C14 correct? – wildpanda Oct 20 '22 at 16:38
  • Please do not deface questions once your use is completed. Future users who have the same or similar questions can still benefit from them. – JMoravitz Oct 20 '22 at 17:04

1 Answers1

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Recapping comments:

Your initial attempt was wrong because you incorrectly applied some significance to whether a person was picked in the initial three for a department (a "first-stringer" if you will) or if they were picked as part of the leftovers. As a result you overcounted.

You ask "should I have added then" If by this you mean adding instead of multiplying, no... that wouldn't fix it. Multiplying is close to correct except you are overcounting. Multiplication principle (rule of product) works specifically when we are only describing each outcome once, not more than once.

To correct this, we approach by way of breaking into cases based on how many English department members were chosen. The number of ways to have $k$ English department members and $14-k$ CS department would be $\binom{15}{k}\binom{13}{14-k}$. Adding up over all possible cases for $k$ which make it so that we have at least three members from each, that would be an answer of:

$$\binom{15}{3}\binom{13}{11}+\binom{15}{4}\binom{13}{10}+\binom{15}{5}\binom{13}{9}+\dots+\binom{15}{11}\binom{13}{3}$$

That is eleven terms being added. It does feel like a lot. If you prefer, you could get away with a bit less arithmetic by approaching indirectly, taking away those teams who fail to have at least three from each from the total if we ignore that... giving an answer of:

$$\binom{28}{14} - \binom{15}{1}\binom{13}{13}-\binom{15}{2}\binom{13}{12}-\binom{15}{12}\binom{13}{2}-\binom{15}{13}\binom{13}{1}-\binom{15}{14}\binom{13}{0}$$

Note, these two expressions are of course equal since they counted the same thing. If you were to move everything but the $\binom{28}{14}$ to the one side, you wind up seeing the identity $\binom{28}{14}=\binom{15}{1}\binom{13}{13}+\binom{15}{2}\binom{13}{12}+\dots+\binom{15}{14}\binom{13}{0}$, which is just an example of Vandermonde's Identity

JMoravitz
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