3) How many positive divisors does the number $5 \cdot 4 \cdot 3 \cdot 2$ have? (Among the divisors of a number, 1 and the number itself must also be counted.)
(A) 4 , (B) 8 ,
(C) 10
(D) 12 ,
(E) 16 .
3. The answer is $(\mathbf{E})$.

The prime factorization of $5 \cdot 4 \cdot 3 \cdot 2$ is: $2^{3} \cdot 3 \cdot 5$. A divisor of this number will therefore have a prime factorization of the form $2^{a} \cdot 3^{b} \cdot 5^{c}$ where $a=0,1,2,3, b=0,1, c=0,1$. 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 4 possible choices for the value of $a$, two for the values of $b$ and of $c$, so in total the possible triplets $(a, b, c)$ are $4 \cdot 2 \cdot 2=16$.