In how many ways can a basketball coach form a team of 2 centers, 2 guards and 1 forward out of a total of 10 basketball players, if
3 players can only play center,
3 only guard position,
1 only forward position,
2 players guard or forward,
1 center or forward?
The solution is as follows: ${5\choose 2}\cdot{4\choose 2}+{5\choose 2}\cdot{3\choose 2}+2\cdot{4\choose 2}\cdot{4\choose 2}-{3\choose 1}\cdot{4\choose 2}=144$
I understand ${5\choose 2}\cdot{4\choose 2}$ is for the case when we picked both "OR" cases as not forwards, but I don't get how we got the rest.
Best Answer
First, pick who the forward is. Break into cases based on which category the forward came from, be it the only forward category, the guard or forward category or the center or forward category.
In the event that your forward you picked was the person who could only play forward, you still have five people who can play guard (the 3 only guard plus the 2 guard or forward) of which we want to pick two of them and then you want to pick two people from the four you still have who can play center (the 3 only center plus the 1 center or forward). This accounts for $\binom{5}{2}\binom{4}{2}$ different configurations.
The other terms in the above is what happens if the forward you selected for your team came from one of the other categories. Supposing they came from the guard or forward category, that gives two options for who it was. Next, pick who the two centers are and who the two guards are. This is the $2\cdot \binom{4}{2}\binom{4}{2}$ term.
Similarly if the forward you select for your team came from the center or forward category, that leads to the $\binom{5}{2}\binom{3}{2}$ term.
I would have stopped here personally. I would have expected the answer to have been $\binom{5}{2}\binom{4}{2}+\binom{5}{2}\binom{3}{2}+2\binom{4}{2}\binom{4}{2}$, having considered a team as being comprised of members who have clearly defined roles.
The $-\binom{3}{1}\binom{4}{2}$ term that appears seems to be trying to correct for something with some unwritten assumption opposite to what I had assumed, so let us explore that in greater detail to see if we can figure out what was intended...
Let the players be denoted by the digits $0,1,2,\dots,9$. Let the only center players be denoted as $OC = \{0,1,2\}$, the only guard as $OG=\{3,4,5\}$, the only forward as $OF=\{6\}$, the guard or forward as $GF=\{7,8\}$ and the center or forward as $CF=\{9\}$
Let us denote a team using the digits and a subscript detailing what position they are on the team to play as.
I would have considered the team $0_C,1_C,3_G,7_G,8_F$ to be a different team than $0_C,1_C,3_G,7_F,8_G$. It is possible that the author of the problem had intended these to be treated both as $0,1,3,7,8$ and are indistinguishable. If that is the case, then let us consider which cases we overcounted. Overcounting will only have occurred if it is unclear who was playing what position.
It is clear that if we have player $6$ on our team (the person who can play only forward) that he must be playing forward, so that case did not lead to any overcounting. If player $6$ was not in our team and it was instead player $9$ who was playing forward, it could not be confused with another team arrangement because switching $9$ to playing center instead there isn't a center to swap back with them to keep the team balance correct and it would have changed it to being $3$ centers.
As such, the only situation we might have overcounted then was if both players $7,8$ were on the team with one of them playing forward. We will have counted it once with $7$ as the forward and $8$ as a guard and again with $8$ as the forward and $7$ as a guard but otherwise the same teammembers. To count how many times this happened, we know $7$ and $8$ are both on the team. We still need to pick one guard and two centers, which can be done in $\binom{3}{1}\binom{4}{2}$ ways.
Since we had counted these twice, subtracting $\binom{3}{1}\binom{4}{2}$ has the effect of making it so that overall they will have been counted exactly once, bringing the final total to:
$$\binom{5}{2}\binom{4}{2}+\binom{5}{2}\binom{3}{2}+2\binom{4}{2}\binom{4}{2}-\binom{3}{1}\binom{4}{2}$$