Question 984204
<pre>
Suppose you have a row of 33+5-1 or 37 o's in a row like this: 

o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o 

Suppose we choose 5-1 or 4 of these to color black, like this:

o o o o o o o &#9679; o o o o &#9679; o o &#9679; o o o o o o o &#9679; o o o o o o o o o o o o o 

There are 7 o's left of the first &#9679;. 
There are 4 o's between the 1st and 2nd &#9679;'s. 
There are 2 o's between the 2nd and 3rd &#9679;'s.
There are 7 o's between the 3rd and 4th &#9679;'s.
There are 13 o's right of the 4th &#9679;.

This represents the solution 7+4+2+7+13 = 33 to x1 + x2 + x3 + x4 + x5 = 33

--------------------------------- 

Suppose we pick these 4 to color black:

o o o o o o o o o o o o o o o o &#9679; &#9679; &#9679; &#9679; o o o o o o o o o o o o o o o o o

There are 16 o's left of the first &#9679;. 
There are 0 o's between the 1st and 2nd &#9679;'s. 
There are 0 o's between the 2nd and 3rd &#9679;'s.
There are 0 o's between the 3rd and 4th &#9679;'s.
There are 17 o's right of the 4th &#9679;.

This represents the solution 16+0+0+0+17 = 33 to x1 + x2 + x3 + x4 + x5 = 33

--------------------------------- 

Or these:

o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o &#9679; &#9679; &#9679; &#9679;

There are 33 o's left of the first &#9679;. 
There are 0 o's between the 1st and 2nd &#9679;'s. 
There are 0 o's between the 2nd and 3rd &#9679;'s.
There are 0 o's between the 3rd and 4th &#9679;'s.
There are 0 o's right of the 4th &#9679;. 

This represents the solution 33+0+0+0+0 = 33 to x1 + x2 + x3 + x4 + x5 = 33

-------------------------------

Or these:

&#9679; o o o o o o o o o o o &#9679; o o o o o o o o o o o &#9679; o o o o o o o o o o o &#9679;

There are 0 o's left of the first &#9679;. 
There are 11 o's between the 1st and 2nd &#9679;'s. 
There are 11 o's between the 2nd and 3rd &#9679;'s.
There are 11 o's between the 3rd and 4th &#9679;'s.
There are 0 o's right of the 4th &#9679;. 

This represents the solution 0+11+11+11+0 = 33 to x1 + x2 + x3 + x4 + x5 = 33

-------------------------------

So we see that any solution of x1 + x2 + x3 + x4 + x5 = 33 can be represented
by coloring 4 of the 37 o's black.

So we have 37 o's choose 4 to color black.

Answer:  37C4 = 66045 

The formula for the number of unordered partitions of n into the sum of
r non-negative integers is {n+r-1)C(r-1) or maybe you write it {{{(matrix(2,1,n+r-1,r-1))}}}.  
This is the case n=33 and r=5

Edwin</pre>