# Integral solutions to $x_1 \cdot x_2 \cdot x_3 \cdot x_4 = 210$

Find how many integral solutions and there to the given condition for $x_1 , x_2 , x_3$ and $x_4$

$$x_1 \cdot x_2 \cdot x_3 \cdot x_4 = 210$$

I factored it to $2 \cdot 7 \cdot 5 \cdot 3$, Then how do I proceed?

• Is the solution $x_1=1, x_2=70,x_3=3,x_4=1$ to be considered different from $x_1=3, x_2=1,x_3=1,x_4=70$? – André Nicolas Jan 27 '15 at 9:59

Since the prime factorization of $$210=2\cdot 3\cdot 5\cdot 7$$
What we have to basically do is find out the number of ways we distribute these factors among the four variables $x_1,x_2,x_3$ and $x_4$.
We can distribute each factor among the $4$ variables by ${4\choose 1}=4$ ways.
$$4^4=256$$
$$({4\choose 0}+{4\choose 2}+{4\choose 4})=8$$
This is because for negative numbers, only $2$ or all $4$ variables have to be negative to obtain positive product.
$$256\times 8=2048$$
• You need $\binom 40$ in there as well to capture the positive solutions! – Mark Bennet Jan 27 '15 at 11:28
• Oh... for some reason I thought that ${4\choose 0}=0$. Silly me. Thanks @MarkBennet – AvZ Jan 27 '15 at 11:30