a. The digits of N! can be rearranged to form a prime.

b. N>2

Find the smallest N, complying with the above statements.

(In reply to

What, no answer yet? by Ady TZIDON)

OK, I'll bite.

3! = 6. Doesn't work.

If N > 3 then N! is divisible by 3, so the sum of its digits is divisible by 3, before and after rearrangement, so any number formed by rearranging its digits is still divisible by 3.

Thus, no qualifying N exists.