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.
Best Answer
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.