n!n! as product of consecutive numbers

Let n be a positive integer.

In how many ways can one write n! as a product of consecutive integers?

For example: 4!=1\times2\times3\times4=2\times3\times4. Here, 2 possibilities exist.

5!=1\times2\times3\times4\times5=2\times3\times4\times5=4\times5\times6. Here, 3 possibilities exist.

Answer

The products of consecutive integers, (m+1)\cdot (m+2)\cdot(m+3)\cdots n is a factorial from which you have removed the first factors, i.e \dfrac{n!}{m!}. To make this equal another factorial, you need to discard the largest factors of n!, i.e. n\cdot(n-1)\cdot(n-2)\cdots.

It is always possible to discard n itself. This yields
\frac{(m!)!}{m!}=(m!-1)!
For instance
5\cdot6\cdots24=1\cdot2\cdots23.
and there are infinitely many solutions, extremely large.
Other solutions are accidental and rare.

Attribution
Source : Link , Question Author : 153 , Answer Author : Community

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