1
$\begingroup$

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: $x_1 \geq 1, x_2 \geq 2, x_3 \geq 3$.

$\endgroup$
  • $\begingroup$ 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
4
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.