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?
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 $C_n^k=\frac{n!}{k!\cdot(n-k)!}$