I know the number of nonnegative integer solutions to the inequality $$x_1 + x_2 + x_3 \leq 10$$ could be computed by adding a slack variable $x_4$ and thus calculating the number as $\binom{13}{3}$. But how would we solve the same problem if we add the following restrictions: $x_1 \geq 1, x_2 \geq 2, x_3 \geq 3$.

  • $\begingroup$ I have corrected the "number of integer solutions" to "number of nonnegative integer solutions" as recommended by Marc Lao. $\endgroup$ – O.A. Jan 17 '15 at 8:34

Hint: Rewrite $x_1 + x_2 + x_3 \leq 10$ as

$(y_1 + 1) + (y_2 + 2) + (y_3 + 3) \leq 10$, where the $y_i$'s are nonnegative integers.

Can you see why this works?

This simplifies to $y_1 + y_2 + y_3 \leq 4$. Then just apply the previous technique you used.

PS: The first sentence in your question should have read

I know that the number of the NONNEGATIVE integer solutions to the following inequality.

Otherwise, there will be infinitely-many solutions.

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