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Question: A basketball team has 5 players, 3 in forward position (which includes a center) and 2 in guard position. In how many ways can we make a team if there are 6 forwards, 4 guards and 2 people who can play forward or guard?

The way I am approaching this question is as follows: We need to pick 3 people out of 6 for the forward position: $\binom{6}{3}$.

We also need to pick 2 guards out of possible 4: $\binom{4}{2}$.

So far, the answer is: $\binom{6}{3} \times \binom{4}{2}$. (assuming that the 2 people who can play forward or guard statement is disregarded).

If two people can play forward or guard, I am thinking of assuming both play forward, add that to assuming one is forward one is guard, plus assuming two are guards. So:

$\binom{8}{3}\binom{4}{2} + \binom{7}{3}\binom{5}{2} + \binom{6}{3}\binom{6}{2}$

Would this yield the right answer? If not, why?

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up vote 4 down vote accepted

I would do this carefully, the long way, and then see whether I have missed any shortcuts. A procedure that does not involve crossing the fingers has high priority!

You counted correctly the number $\binom{6}{3}\binom{4}{2}$ of teams made up of specialists. So there are $120$ such teams. Now we count separately the teams that contain $1$ versatile player, and $2$ versatile players.

If we are going to have $1$ versatile player, she can be chosen in $\binom{2}{1}$ ways. She can replace a forward, leaving $\binom{6}{2}\binom{4}{2}$ choices for the rest of the team, or she can replace a guard, leaving $\binom{6}{3}\binom{4}{1}$ choices for the rest of the team. Thus there are $$\binom{2}{1}\left(\binom{6}{2}\binom{4}{2}+\binom{6}{3}\binom{4}{1}\right)$$ teams with exactly $1$ versatile player. So there are $340$ such teams.

If we are going to use $2$ versatile players, they can be both replace forwards, leaving $\binom{6}{1}\binom{4}{2}$ choices, or both replace guards, leaving $\binom{6}{3}$ choices, or do one of each, leaving $\binom{6}{2}\binom{4}{1}$ choices, for a total of $$\binom{6}{1}\binom{4}{2}+\binom{6}{3}+\binom{6}{2}\binom{4}{1}$$ teams with exactly $2$ versatile players. So there are $96$ such teams.

Finally, add up.

Remark: I interpreted team to mean a set of $5$ people. If by team we mean a set of $5$, together with a specification of what positions (forward or guard) they are playing, the answer would be different, since one versatile player in each position would have to be counted twice, once for X playing forward and Y playing guard, and once for the reverse. And one can complicate things further.

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I like this approach. Would: $\binom{8}{3}\binom{4}{2} + \binom{7}{3}\binom{5}{2} + \binom{6}{3}\binom{6}{2} + \binom{6}{3}\binom{4}{2}$ be equivalent? This assumes both are forwards, then 1 is forward and 1 is guard, then 2 are guards, then none. If not, would would this be wrong? – darksky Apr 10 '12 at 18:56
I believe the question does not require those two players to play. Your final answer assumes both DO play. Correct? – darksky Apr 10 '12 at 18:58
@Nayefc: I produced three numbers, which need to be added. Do not trust my typing (I am typo-prone) or my multiplications (I am error-prone) but the reasoning is right, and more importantly, safe. – André Nicolas Apr 10 '12 at 19:06
So the problem with my above comment over counts. True? – darksky Apr 10 '12 at 19:07
@AndréNicolas Your very last term should be $2 \binom{6}{2}\binom{4}{1}$, since "one of each" could mean the first player is a guard and the second a forward, or vice versa. Making this change causes the total to agree with mine. – Austin Mohr Apr 10 '12 at 19:08

Note: For this answer, I interpret "team" to mean a set of five players together with their position. Thus, in my counts, it makes a difference whether a versatile player is playing guard or forward.

If you just lump your flexible players into each category, you have 8 forwards and 6 guards. So, a rough estimate for the number of teams is $\binom{8}{3}\binom{6}{2}$. The problem with this is some configurations feature the same person in two positions, which is unacceptable. If we subtract those, we'll have the correct count.

Let $A$ be the set of teams featuring the first flexible player twice and let $B$ be the set of teams featuring the second flexible player twice. We want to find $|A \cup B|$, which is $|A| + |B| - |A \cap B|$ by inclusion-exclusion.

Now, $|A| = \binom{7}{2}\binom{5}{1}$, since we must fill out the team under the assumption that the first player is already playing as both guard and forward. Similarly, for $|B|$. Finally, $|A \cap B| = \binom{6}{1}$, since both the flexible players are playing forward and guard.

Gathering everything up, we have the number of teams being $$ \begin{align*} \binom{8}{3}\binom{6}{2} - |A \cup B| &= \binom{8}{3}\binom{6}{2} - \left(2\binom{7}{2}\binom{5}{1} - \binom{6}{1}\right)\\ &= 636. \end{align*} $$

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When you write $\binom{8}{3}\binom{4}{2} + \binom{7}{3}\binom{5}{2} + \binom{6}{3}\binom{6}{2}$, you consider the possibility of not using the free positioned player at all, thrice. So, I think you should subtract $2\times \binom{6}{3}\times \binom{4}{2}$ from $\binom{8}{3}\binom{4}{2} + \binom{7}{3}\binom{5}{2} + \binom{6}{3}\binom{6}{2}$.

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what does $2\times \binom{6}{3}\times \binom{4}{2}$ correspond to? – darksky Apr 10 '12 at 18:44
you know, the number of ways of not using the free positioned player at all. You considered it thrice in your expression, when you should have to consider it only once. So subtract 2 times of that. Helps? – funktor Apr 10 '12 at 18:50
Would: $\binom{8}{3}\binom{4}{2} + \binom{7}{3}\binom{5}{2} + \binom{6}{3}\binom{6}{2} + \binom{6}{3}\binom{4}{2}$ be equivalent? This would include not using him at all once. – darksky Apr 10 '12 at 18:54
may be not. they yield different results, you're not adding, you are to subtract – funktor Apr 10 '12 at 18:58
Yes, but when you have say 8 possibility of forwards, you have the liberty to chose just from the original 6 forwards too – funktor Apr 10 '12 at 19:02

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