Different possible integer solutions for the given function

Question

How many integer solutions exist for the given equation $x+y+z=15$ subject to the constraint that $0\leq x,y,z\leq 10$?

I tried the brute force method and listed the possible solution sets for the above equation, given the variable constraints as follows :

$\begin{bmatrix} 2 &3 &10 \\ 2&4 &9 \\ 2&5 &8 \\ 3&6 &6 \\ 3&4 &8 \\ 3&5 &7 \\ 4&5 &6 \\ 4&7 &4 \\ 4&10 &1 \\ 4&11 &0 \\ 6&7 &2 \\ 6&8 &1 \\ 6&9 &0 \\ 5&9 &1 \\ 5&10 &0 \\ 7&8 &0 \end{bmatrix}$

Here, each row will denote the possible values for the three variables. I tried to eliminate any duplicate values, and listed only the unique combinations of integer values. Then, the number of possible solutions should be one of the permutations of the above values only, right? So, that should be 16 * 6 = 96 possible solutions. The answer given was 3666, derived using Generating functions. Please explain.

Comments

Take x = 3, y = 6, z = 6. Now in your 6 permutation you will repeat these values rt?

---

Yes. That was the idea. To find the uniques combinations and then get permutations to each solution. That should be the total number of solutions possible.

Answer 1

This problem can be reduced to balls in bins problem. $x$, $y$ and $z$ are 3 distinct bins and we have 15 identical balls to fill them. Now, the problem reduces to all the possible permutations of 15 + (3-1) (3-1 separations split 15 items to 3 bins) items, with 15 identical balls and another 2 identical separations. So,

$$\frac{17!}{15!2!} = 136.$$

Now, we have to eliminate the cases where $x,y,z > 10$.

So, let $x = 11$, now we have $y + z = 4$, which has 5 possible cases (calculated as above). Similarly for $y=11$ and $z=11$ we get 5 more cases each totaling 15 cases.

Similarly for $12, 13, 14, 15$ we get $3.4, 3.3,,3.2, 3.1$ cases totaling 15 + 12 + 9 + 6 + 3 = 45 possible cases.

So, our answer should be $136 - 45 = 91.$

Comments

Sir, 17C2 will be 136. So, the answer should be 91, right?

---

Thanks. corrected. In your approach you can add [5,5,5] and remove 3 cases each for [3,6,6] and [4,7,4] to get 91 :)

---

@Arjun  $sir$ is this also a right approach using GF

$[x^{15}]\ in\ (1+x+x^2+x^3..+x^{10})*(1+x+x^2+x^3..+x^{10})*(1+x+x^2+x^3..+x^{10})$

$[x^{15}]\ in\ \left(\dfrac{1-x^{11}}{1-x}\right)^3$

$[x^{15}]\ in\ \sum_{k=0}^{3}\ ^3C_k(-x^{11})^k\times \sum_{k=0}^{\infty}\ ^{2+k}C_k(x^k)$

$=\ ^{17}C_{15}-\ ^{3}C_{1}.\ ^{6}C_{4}$

$=136-45$

$=91$