8) How many positive divisors does $6 !=1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6$ have? (Among the divisors of a number, 1 and the number itself must also be counted.)
(A) 5
(B) (C) 10 ,
(D) 24 ,
(E) 30 .
8. The answer is $(\mathbf{E})$.

The prime factorization of $6 !=6 \cdot 5 \cdot 4 \cdot 3 \cdot 2$ is: $2^{4} \cdot 3^{2} \cdot 5$. A divisor of it must therefore have a prime factorization of the form $2^{a} \cdot 3^{b} \cdot 5^{c}$ where $a=0,1,2,3,4, b=0,1,2$, $c=0,1$. Each divisor is identified by a specific choice of the triplet $(a, b, c)$. Counting the divisors is equivalent to counting how many distinct triplets can be formed. There are 5 possible choices for the value of $a$, three for the value of $b$ and two for the value of $c$, so in total the possible triplets $(a, b, c)$ are $5 \cdot 3 \cdot 2=30$.