Let $N$ be the number of ways of distributing $52$ identical balls into $4$ distinguishable boxes such that no box is empty and the difference between the number of balls in any two of the boxes is not a multiple of $6$. If $N = 100a+b$, where $a,b$ are positive integers less than $100$, find $a+b$.
Thanks in Adv for the Help !
Till now i have tried with ${{n-1}\choose{r-1}} = \frac{(n-1)!}{(r-1)!(n-r)!}$, ( Combination formula ), where $n$ is no of balls and $r$ is the no of boxes, With this I got $425$ as $N$
But i am unable to get to understand how to change this to adapt the questions imp. points
- $6 \not | $ diff between $2$ box
- $N = 100a+b$