Find the value(s) of $n$ that satisfy the equation $P(n, 3) = 8n + P(n, 2)$ where $P(n,k)=\frac{n!}{(n-k)!}$.
So far I have : $$P(n,3) = 8n + P(n, 2) =8n + n(n-1) =7n + n^2 $$
How do I expand $P(n,3)$?
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Dave
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lollol12345
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1We have that: $$P(n,3) = \frac{n!}{(n-3)!} = \frac{n(n-1)(n-2)(n-3)!}{(n-3)!} = n(n-1)(n-2)$$ – Mark Schultz-Wu Sep 03 '17 at 00:50
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Well, you have $$P(n,3)=\frac{n!}{(n-3)!}=\frac{n(n-1)(n-2)(n-3)!}{(n-3)!}=n(n-1)(n-2)$$ and $$P(n,2)=\frac{n!}{(n-2)!}=\frac{n(n-1)(n-2)!}{(n-2)!}=n(n-1)$$ so we are trying to solve $$n^3-3n^2+2n=8n+n^2-n$$ or equivalently $$n^3-4n^2-5n=0$$
Dave
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