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How many integral values of n exist such that $n>2$ and $\frac{n!}{(n-2)!} <150$?

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    n! = n*(n-1)*(n-2)! – 2017-01-25
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    Do you mean $\frac{n!}{n-2!}$ or $\frac{n!}{\color{red} (n-2\color{red} ) !}$ ? – 2017-01-25

2 Answers 2

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We have $$\frac {n!}{(n-2)!} =\frac {(n-2)!(n-1)(n)}{(n-2)!} = (n-1)(n)= 2\binom {n}{2} $$ We need $2\binom {n}{2}<150 \Rightarrow \binom {n}{2}<75$. Also we can easily check that $\binom {12}{2} =66$ and $\binom {13}{2}=78$. So the values of $n $ satisfying the condition are: $$n\in [3,12] \in \mathbb N$$ Hope it helps.

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    how do the binomial coefficients help? – 2017-01-25
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    @JorgeFernándezHidalgo I just thought giving it a combinatorial flavor would like nice. – 2017-01-25
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$\frac{n!}{(n-2)!}=n^2-n$.

This function is increasing as $(n+1)^2-(n+1)-(n^2-n^2)=2n$.

Notice that the first few values for $n>2$ are: $6,12,20,30,42,56,72,90,110,132,156$. So the answer is: $n=3,4,5,6,7,8,9,10,11,12$.