How to find the largest integer $n$ for which $n!$ can be expressed as the product of $n - 3$ consecutive integers

Question

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 $

Answer 1

Notice that $(n+1)\times n \times \dots \times 5 > n!$ for $n\geq 24$. So we must have $n\leq 23$.

On the other hand $(23+1)\times 23 \times \dots \times 5 = 23!$ . So the answer is $23$, like the Miley Cyrus song.

Answer 2

$n=23$ 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 $n\leq 23$.

Answer 3

Clearly $n!>n!/3!$. Now $$ n!<(n+1)!/4! \Leftrightarrow 24=4!<n+1, $$ i.e., $n>23$. Therefore it is enough to check manually the cases $n\le 23$.

$\bullet$ $n!=(n+1)!/4!$ has solution $n=23$. This is obviously the largest.