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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?

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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.

Hence, the answer is

$$4^4=256$$

Also considering negative numbers, we can multiply the result by

$$({4\choose 0}+{4\choose 2}+{4\choose 4})=8$$

This is because for negative numbers, only $2$ or all $4$ variables have to be negative to obtain positive product.

So, the final answer will be

$$256\times 8=2048$$

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