P and C:--

For positive non-primes $m \leq 4$ , we have $1\mid (1-1)!=1, 4\nmid(4-1)!=6$ . So the only such non-prime value $\leq 4$ fulfilling the conditions in the question is $4$ .

For non-primes $m > 4$ , we can express $m$ as a product $m=d_1*d_2$ , where $d_1, d_2$ are distinct proper factors of $m$ (proper factors being factors of the number not equal to 1 or the number itself). We have $d_1, d_2$ being factors of $(m-1)!$ , since $d_1,d_2\leq m-1$ and since the two are distinct, they are two non-repeating terms in the product $(m-1)!=1 \times 2 \times \dots \times (m-1)$ , hence we have $m=d_1*d_2 \mid (m-1)!$ for all non-primes $m>4$

In conclusion, the only non-primes $m$ that satisfy this problem's conditions are $m=4$ , consequently the number of such non-primes is $\boxed{1}$

(m-1)! = (m-1)(m-2)(m-3)(m-4)......................3.2.1 For all prime values of m and 4, (m-1)! is not divisible by 'm' as for 'm'=4,,,3!/4=3/2 i.e it is not exactly divisible. Hence the answer is '1'.