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How many integer solutions of $x_1+x_2+x_3+x_4=28$ are there with $-10\leq x_i\leq20$?

I know how to do this type of problem when all $x_i$ are positive, but I am struggling to understand how to complete this problem when the $x_i$ can be negative.

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marked as duplicate by N. F. Taussig combinatorics Oct 31 '18 at 12:34

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    $\begingroup$ Try defining $y_i = x_i + 10$ $\endgroup$ – Sambo Oct 31 '18 at 3:30
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Guide:

Let's see, you can manage the question, when all $x_i$ are positive.

let $y_i = x_i + 11$, then we have $1 \le y_i \le 31$.

and $$\sum_{i=1}^4x_i = 28$$

would become $$\sum_{i=1}^4y_i= 28 + 4(11)$$

$1 \le y_i \le 31$.

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  • $\begingroup$ Thanks! This was really helpful! I think I understand it now. $\endgroup$ – lmckal45 Oct 31 '18 at 3:34
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This is a nice question. You know how to solve these problems with stars and bars, I gather. So we want to know how many ways there are to put $28$ indistinguishable balls in $4$ distinct buckets, with at least $-10$ and no more than $20$ balls in each bucket. Start out by putting $-10$ balls in each bucket -- difficult in the real world, but no trouble at all in math. Now we have $-40$ balls, so to get up to $28$ we need to add $68$ balls to the buckets. Since we aren't allowed to have more than $20$ balls in a bucket, we can't place any more than $30$ balls in a bucket.

So the question is transformed into, "In how many ways can $68$ indistinguishable balls be placed in $4$ distinct buckets, with no more than $30$ balls in a single bucket?"

Take it from here.

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Find out how many integer solutions are for $u_1+u_2+u_3+u_4=40+28$ with $0\leq u_i \leq 10+20$. Then associate each $u_i$ with $x_i=u_i-10$.

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