.
This problem and the method of its solution are of the highest peaks in Combinatorics.
They are at the level of a School Math circle at the local or (better to say) a renowned University.
x1 + x2 + x3 + x4 + x5 = 36 (1)
Imagine 36 marbles on the table, placed in one straight line with small gaps between them.
Imagine you have 6 numbered bars, of which you place the first bar (bar N1) before the first marble and the last bar (bar N6)
after the last marble in the row.
Let , , , , be some solution to the given equation.
You then place bar N2 after -th marble in the gap in the row of marbles;
then you count next marbles in the row of marbles after bar N2 and place bar N3 in the gap there;
then you count next marbles in the row of marbles after bar N3 and place bar N4 in the gap there;
finally, you count next marbles in the row of marbles after bar N4 and place bar N5 in the gap there.
At this moment, all 36 marbles are divided in 5 groups between bars (1-2), (2-3), (3-4), (4,5) and (5,6).
Notice that if some is zero, then the corresponding bars go to common respective gap.
So, having the solution to equation (1) in non-negative positive integer number, you place 4 bars N2, N3, N4 and N5 in
their corresponding positions in gaps in the row of marbles.
Vise versa, if you place 4 bars B2, B3, B4 and B5 in gaps in the row of 36 marbles, you divide marbles in 5 groups,
and the numbers of marbles in each group form the solution to equation (1).
Thus, there is one-to-one correspondence between the set of solutions to equation (1) in non-negative integer numbers,
from one side, and all different possible placings of 4 bars in 35 gaps in the row between 36 marbles.
Thus we have 35+4 = 39 entities, 35 marbles and 4 bars; 35 marbles are indistiguishable and 4 movable bars
are indistinguishable, too.
The number of all possible indistinguishable arrangements of 39 items of two types with 35 indistinguishable of one type
and 4 indistinguishable of the other type is = = 82251.
Hence, the number of all possible solutions to equation (1) in non-negative integer numbers is equal to = 82251.
ANSWER. The number of different solutions to equation (1) is 82251.
Solved.
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More general problem to find the number of non-negative integer solution to equation
+ + + . . . + = n
where k <= n, can be solved in the same way and has the answer .
We have then (n-1) gaps between "n" "marbles" and (k-1) movable dividing bars.
The method I used in the solution is called the "method of bars and stars".
You can read about it in this Wikipedia article
https://en.wikipedia.org/wiki/Stars_and_bars_%28combinatorics%29