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In an alternate universe, LA Lakers decided to gather the 5 best guards and 10 best forward/centers in their history. One of the guards is Kobe, and one of the forward/centers is Shaq. How many 5-man starting lineups are possible consists of 2 guards and 3 forward/centers if Kobe and Shaq refuse to be starters at the same time?

Why is $504$ wrong?

I solved both of them using: $\dfrac{n!}{r!(n - r)!}$

And got $6 × 84 = 504$

I $n - 1$ both of them due to the condition of the problem. Is there something wrong with my solution which makes $504$ wrong?

X X
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    You seem to have done ${4 \choose 2}{9 \choose 3}$ which is the number if neither Kobe nor Shaq start. But what happens if one of them starts? – Henry Mar 27 '20 at 20:37
  • Do I have to find the permutation where they are both included? Sorry I am very confuse – X X Mar 27 '20 at 20:41
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    You have to find the number of line-ups where neither is included plus the number of line-ups where only one of the two is included (since they refuse to play together). It might be easier to find the total number of line-ups with no restrictions and then subtract the number of line-ups where both are included. – NickD Mar 27 '20 at 20:44
  • I got 504 + 840 + 720, is this right? – X X Mar 27 '20 at 20:51

1 Answers1

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You should find the number of all line-ups and then subtract the number of those where both of them are starters at the same time: $$C_5^2 \cdot C_{10}^3-C_4^1\cdot C_9^2=1200-144=1056$$ Where $C_n^k=\frac{n!}{k!\cdot(n-k)!}$

VIVID
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