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Thread: Combinatorics question

  1. #1
    Jan 2013

    Combinatorics question

    So you have 12 players paul can play all 3 positions, 5 can only play center or
    forward and 6 can only play guard. How many basketball lineups can be made? (there are 1 center 2 guards and 2 forwards and the order of guards and forwards doesn't matter)

    we have c1,f1,f2,g1,g2 are the positions I broke it into cases paul
    playing c1,f1... No position and then use addition rule to get all lineups

    So paul plays c1 = (1*5*4*6*5)/(2*2) I divided by 4 since the order of
    the forwards and guards doesn't matter

    Paul plays f1= ( 5*1*4*6*5)/2 I
    divided by 2 since now the forward position is fixed but still the guard
    position doesn't matter

    And the others follow same idea is this correct? Thank you
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  2. #2
    Senior Member
    Mar 2008
    Pennsylvania, USA

    Re: Combinatorics question

    Consider the possible lineups when Paul is the center:
    You have six guards and from which you must choose two. There are $\displaystyle \binom{6}{2}=15$ ways to select the guards.
    Since Paul is playing center, you lastly need to fill up the 2 forward positions choosing from a pool of 5. There are $\displaystyle \binom{5}{2}=10$ different guard pairs that can be in your starting lineup. So, there are $\displaystyle 15\times10=150$ possible lineups with Paul serving as the starting center.

    Next, consider Paul plays forward:
    In this case, let's select two from the pool of 5 hybrid forward/centers to consist of a (center,forward) pair. There are $\displaystyle \binom{5}{2}=10$ possible pairings to play center and forward, without regard to whom are which. Now, since who plays center and who plays forward does matter in this problem, we can multiply the 10 pairings by 2 to determine the number of possible assignments to the center and forward, when Paul starts at forward. So, when Paul starts at forward, there are $\displaystyle 10\times2=20$ possible ways in which the forwards and centers can be assigned as starters. Finally, the the number of ways to assign guards to the starting lineup is given by $\displaystyle \binom{6}{2}=15$. So, when Paul starts at forward, there are $\displaystyle 20\times15=300$ different possible lineups.

    Now, we consider when Paul starts as a guard:
    In this circumstance, there are $\displaystyle \dbinom{6}{1}=6$ ways to choose the other guard. There are $\displaystyle \binom{5}{3}=10$ ways (10 unique triplets) to select the remaining starters (center, forward, forward). For each of these 10 ways -- or unique triplets -- there are 3 different ways to assign the center, i.e., (c,f,f), (f,c,f), (f,f,c). So, there are $\displaystyle 3\times10=30$ ways to assign the frontcourt (the center and forwards). Thus, when Paul starts at guard, there are $\displaystyle 6\times30=180$ possible lineups.

    Finally, we consider it when Paul is not a starter:
    There are $\displaystyle \binom{6}{2}=15$ ways to select the guards. To select the final three starters, there are $\displaystyle \binom{5}{3}=10$ possible ways. Again, for each forward-forward-center triplet, there are 3 possible ways to assign the center. So, the forwards and centers can be selected in $\displaystyle 3\times10=30$ different ways. So, when Paul does not start, there are $\displaystyle 15\times30=450$ ways to select the lineup.

    Adding these up, we have $\displaystyle 150+300+180+450 = 1080$ possible starting lineups.
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