TITLE: In how many ways can 5 numbers be selected from the numbers 1,2,…,500 so that their sum is a multiple of 5?
QUESTION [6 upvotes]: We have 5 different subgroups of the 500 numbers, where each number has a specific remainder 0 to 4 mod 5.
Each subgroup consists of 100 numbers.
So we must select all 5 numbers from each of the below remainders sets:
(0,0,0,0,0), (1,1,1,1,1), (2,2,2,2,2), (3,3,3,3,3) and (4,4,4,4,4).
But apart from the above, we also have (1,1,1,2,0), (1,2,2,0,0), (1,2,3,4,0), (2,2,2,4,0) etc.
All in all, we have the below:
(0,0,0,0,0)
(0,0,0,1,4)
(0,0,0,2,3)
(0,0,1,1,3)
(0,0,1,2,2)
(0,0,2,4,4)
(0,0,3,3,4)
(0,1,1,1,2)
(0,1,1,4,4)
(0,1,2,3,4)
(0,1,3,3,3)
(0,2,2,2,4)
(0,2,2,3,3)
(0,3,4,4,4)
(1,1,1,1,1)
(1,1,1,3,4)
(1,1,2,2,4)
(1,1,2,3,3)
(1,2,2,2,3)
(1,2,4,4,4)
(1,3,3,4,4)
(2,2,2,2,2)
(2,2,3,4,4)
(2,3,3,3,4)
(3,3,3,3,3)
(4,4,4,4,4)
So the total number of ways is:
$\frac {1} {120}\times\binom {500}1 \times \binom {400}1 \times\binom {300}1 \times\binom {200}1  \times\binom {100}1$ but this is only for the case where we select all 5 numbers from the same subset of remainders. 
How do we add the rest of the cases?

REPLY [2 votes]: The                           following                           MSE
link   asks  about
the  probability that  a set  of size  $n$ drawn  from $[kn]$  has sum
divisible by $n$. We proved the following closed form:
$$\bbox[5px,border:2px solid #00A000]{
(-1)^n
{kn\choose n}^{-1} \frac{1}{n}
\sum_{d|n} {kd\choose d}
(-1)^d \varphi(n/d).}$$
In the present case we have $n=5$ and $k=100$ which gives
$${500\choose 5}^{-1} 51 \; 048 \; 937 \; 600 =
\frac{127622344}{638111719}.$$