<s>[INST] How to find the largest integer $n$ for which $n!$ can be expressed as the product of $n - 3$ consecutive integers I need to find the largest integer $n$ for which $n!$ can be expressed as the product of $n - 3$ consecutive integers. Example: $7! = 7 \cdot 8 \cdot 9 \cdot 10 $
 [/INST] $n=23$ has an answer.
Given that $\binom{K}{n-3} = \frac{K(K-1)\cdots(K-(n-2))}{(n-3)!}$ this means you want a $K$:
$$\binom{K}{n-3} = n(n-1)(n-2)$$
Now, $\binom{K}{n-3}$ is an increasing function as $K$ increases.
And we have:
$\binom{n}{n-3} = \frac{n(n-1)(n-2)}{6} < n(n-1)(n-2)$ and $\binom{n+1}{n-3} = \frac{(n+1)n(n-1)(n-2)}{24}> n(n-1)(n-2)$ when $n+1>24$. So you only have to check  the cases $n\leq 23$.
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