**119**

Largest positive integer n for which n! can be expressed as the product of n-a consecutive positive integers is **(a+1)! - 1**.

**Proof** - let the largest of the n-a consecutive positive integers be k. Clearly k cannot be less than or equal to n, else the product of n-a consecutive positive integers will be less than n!

Now, observe that for n to be maximum the smallest number (or starting number) of the n-a consecutive positive integers must be minimum, implying that k needs to be minimum. But the least k > n is n+1.

So the n-a consecutive positive integers are a+2, a+3, … n+1

So we have

(n+1)!/(a+1)! = n! => n+1 = (a+1)! => n = (a+1)! -1

for a=4 we have n= 5!-1=**119**