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Original Question:

How many ways are there to pick a five-person basketball team from $12$ possible players? How many selections include the weakest and the strongest players?

Solution:

First Part: The order does not matter, so use combinations. $12\choose 5$ or $\frac{12!}{5!(12-5)!}$

Second Part: $10\choose 3$ or $\frac{10!}{3!(10-3)!}$

Personal Logical Question:

For the second part, we have $10$ players to choose $3$ players from because we already have chosen our strongest and weakest player out of the $12$ players we first had. Thus, if a player is not the strongest and weakest, he must be part of the remaining $10$ players. Is my logic right ?

Please provide some clarification.

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    Yes, your logic is right. – 2017-02-16
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    @avs , thank you for confirming. The second part of the question is worded weird. It feels like it is asking how many of the 12 players are strongest and weakest. I guess, it could have been better if the question was worded like: "Considering the strongest and weakest players are already in the starting lineup, how many selections can we have for the remaining spots?". I hope they change the question in the future editions. – 2017-02-16
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    The main shortcoming of the wording is the failure to tell us that there is a *single* strongest player and a *single* weakest one. The rest of it may be difficult to parse, but in real life most questions are like that, if not worse. (As they say in consulting, clients "don't know that they don't know" what they want.:) – 2017-02-16
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    There is also the frustration of not knowing from the wording of the question whether the positions on the court have any impact on our count, whether you can have multiple "*different*" teams using the same five people just by changing which position each person plays (*point guard, power forward, etc...*) – 2017-02-16
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    Yes these kinds of mathematical interruptions are really common especially in discrete math. The best way to answer these kinds of question on a final or midterm would be to just ask the prof to make sure what the question is exactly asking about. – 2017-02-16

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Your logic is spot on, and your solution correct.

There's ... not much more to be said.

edit: Wait.   This is assuming that "the strongest" and "the weakest" are individuals, rather than groups.

If the 12 players are subdivided into 6 strongest and 6 weakest, then the number of ways to select a five person team including players from both subsets is counted by excluding the complement: we take the total count then subtract the count of ways to pick all players just from either category.

$$\binom{12}{5}- 2\;\binom{6}{5}$$

And such.

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    Thank you for confirming. I just wanted to make sure my reasoning is right. – 2017-02-16
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    Just to make sure, are we talking about the second part of the question ? If yes, then I believe your approach "excluding the complement" might be error prone, because the solution provided by the teacher is C(10,3) = 120 – 2017-02-16