Positive integer solution to equation $(x_1+x_2+x_3)(y_1+y_2+y_3+y_4)=15$ What is the total number of positive integer solution to the equation
$(x_1+x_2+x_3)(y_1+y_2+y_3+y_4)=15$
a) 20 $\qquad$ $\qquad$ $\qquad$ $\qquad$ b) 18
c) 10 $\qquad$ $\qquad$ $\qquad$ $\qquad$  d) 4
Sol. $(1,1,1,)(2,1,1,1)$,  $(1,1,1,)(1,2,1,1)$
$(1,1,1,)(1,1,2,1)$ ,  $(1,1,1,)(1,1,1,2)$
Hence there are only 4 positive integer solution to the equation.
Is there any other way to approach this question or this is only one which i have done?
 A: Since $15 = 1\cdot15 = 3 \cdot 5$, the only possibilities are
$$(x_1 + x_2 + x_3, y_1 + y_2 + y_3 + y_4) = (1, 15), (15, 1), (3, 5), (5, 3)$$
Since the variables are all at least $1$, we can rule out the case where $x_1 + x_2 + x_3 = 1$ or where $y_1 + y_2 + y_3 + y_4 = 1, 3$ for the minimum values of each sum exceeds the respective assignments. 
This leaves us with
$$(x_1 + x_2 + x_3, y_1 + y_2 + y_3 + y_4) = (3, 5)$$
Here we see that the only possible assignment for $x_1, x_2, x_3$ is $x_1 = x_2 = x_3 = 1$. So we only need to count the number of solutions to
$$y_1 + y_2 + y_3 + y_4 = 5$$
which is possible via the Stars and Bars method. In fact, you don't even need this at all if you consider the fact that $y_1, y_2, y_3, y_4 \ge 1$.
A: $15 = 5 * 3$, So $(x_1 + x_2  + x_3) = 3$ or $5$ same for the other one.
$(y_1 + y_2  + y_3 + y_4) = 3$ has no solution. $\therefore$  $(x_1 + x_2  + x_3) = 3$ and $(y_1 + y_2  + y_3 + y_4) = 5$. Now you know $x_1, x_2, x_3$ can't be greater than $1$. $\therefore$ the only solution for it is $(1 + 1 + 1)$. In   $(y_1 + y_2  + y_3 + y_4) = 5$, only one variable can be $2$, rest have to $1$. Now we can use permutation here as we have to arrange $2$ in $4$ possible position. 
$$P = {n!\over (n-r)!}$$
$$P = {4!\over (4-1)!} = 4$$  
A: Since all the integers are positive, we can only have
$$x_1+x_2+x_3=3$$
$$y_1+y_2+y_3+y_4=5$$
In the first equation, clearly $x_1=x_2=x_3=1$
In the second,only one is equal to $2$ and everything else is equal to 1
Hence, the solutions:
$$(1,1,1)(2,1,1,1)$$
$$ (1,1,1)(1,2,1,1)$$
$$(1,1,1)(1,1,2,1)$$
$$(1,1,1)(1,1,1,2)$$
4 possibilities.
