Finding number of solutions in the positive integers of (i) $x_1x_2x_3x_4x_5 = 1260$ (ii) $2x + 3y + 4z = 24$ I want to find out the positive integer solutions of variables of the following  : 
(i)  $x_1x_2x_3x_4x_5 =1260$
( ii )  $2x+3y+4z = 24 $ 
MY WORK:
For (i), I only know to find out the factors of $1260$, like $1260=2^2.3^2.5.7$ . So total factors are $36$ . Then,  I try to take combination $\binom{36}{5}$ , for which I get $376992$ . But the answer is $5625$. I know I'm wrong, but can't advance forward.  
For (ii), I am more clueless.  I assume that any number can't be greater than $\frac{24}{4}=6$ ... Then I try to select $3$ values out of $6$ ... For which I get $20$ . But the answer is $19$ .
N. B:  Sorry if I sound silly... I've faced such questions in simple forms earlier only... 
 A: In Part (ii), you cannot actually put $z= 6$ because in that case you'll have to put $x= y = 0$. Thus $z$ can only be from the set $\{ 1, 2, 3, 4, 5 \}$. 
Using the same logic, $y$ can only be from the set $\{ 1, 2, 3, 4, 5, 6, 7 \}$, and $z$ can only be from the set $\{ 1, 2, \ldots, 11 \}$. 
Moreover, since 
$$ 3y = 24 - 4z-2x = 2 ( 12 - 2z-x), $$
so your $y$ must be even. Thus $y$ can only be from the set $\{ 2, 4, 6 \}$. 
Now as $$2(x + 2z) = 2x + 4z = 24 - 3y = 3(8 - y),$$ so $x + 2z$ must be a multiple of $3$. 
Thus $x \in \{ 1, 2, \ldots, 11 \}$, $z \in \{ 1, 2, 3, 4, 5 \}$, and $x+2z$ is a multiple of $3$. 
Thus for $x= 1$, $z \in \{ 1, 4 \}$; for $x= 2$, $z \in \{ 2, 5 \}$; for $x= 3$, $z \in \{ 3 \}$; for $x = 4$, $z \in \{ 1, 4 \}$; for $x= 5$, $z \in \{ 2, 5 \}$; for $x = 6$, $z \in \{ 3 \}$; for $x = 7$, $z \in \{ 1, 4, \}$; for $x= 8$, $z \in \{ 2, 5 \}$; for $x= 9$, $z \in \{ 3 \}$; for $x= 10$, $z \in \{ 1, 4 \}$; and for $x= 11$, $z \in \{ 2, 5 \}$. 
This reasoning narrows down your possibilities considerably. 
Hope this helps. 
