no of possible integral solutions of the problem ??? $xyz=3000$, $x,y,z$ are integers.
Find number of solutions possible.
ATTEMPT: $$3000=2^3\cdot5^3\cdot3$$
                     So computing solution for respective powers independently and multiplying them. That is $$\binom{5}{2}\binom{5}{2}\binom{3}{2} = 300, \binom{n+r-1}{r-1}$$ But the answer given is $1200$. Please help whether I am conceptually wrong or the answer is.
 A: Consider three boxes called $x$, $y$ and $z$.
You have three balls with a $2$, three balls with a $5$ and one ball with a $3$.
There are $\binom 53$ ways to put the $2$-balls in the boxes. There are $\binom 53$ ways to put the $5$-balls, and $3$ ways to put the $3$-ball.
That makes $10\times10\times3=300$.
Now, you have $2$ balls with the sign $-$. There is no need for more, since there are only $3$ boxes and there must be an even number of $-$ balls. There are $3$ ways to put them in different boxes. Putting them in the same box is only one way in this problem. Total: $4$ ways.
$4\times300=1200$.
A: Let $\tau_k(n)$ - number of positive integer solutions of equation $x_1x_2\dots x_k=n$. Then $\tau_k(n)=\sum \limits_{d\mid n}\tau_{k-1}(d)$. In your case you must put $k=3$ and $n=3000$ and you get $\tau_3(3000)=\sum \limits_{d\mid 3000}\tau_{2}(d)$ but $\tau_2(d)=\tau(d)$ - numbers of divisors $d$. If $d=p_1^{\alpha_1}\dots p_m^{\alpha_m}$ then $\tau(d)=\prod \limits_{i=1}^{m}(\alpha_i+1)$
A: $3000=2^3*5^3*3^1$, the total number of factors is one added to each exponent and multiplying the result together i.e $(3+1)*(3+1)*(1+1)=32$, the problem is now to find the number of ways of selecting three factors out of thirty-two i.e $32C_3$
