How to distribute these balls?

Using the stars and bars method, we essentially have an unknown number of balls, $n$ :

$* * * * * * * * *$

and we have to distribute $3$ bars (which act as separators between the jars) in between $(n - 1)$ spaces (because there cannot be empty jars).

An example would be like so:

$*|* *|* * * *|* *$

Therefore, $(n - 1) \text{ choose } 3 = 56 \text{, so } n = 9$ .

$\left(\left(m,n\right)\right)=\binom{n+k-1}{k-1}$

$\left(\left(5,4\right)\right)=56$

Therefore, one for each jar plus five extras for a total of nine.

See also multiset number or stars and bars, which is not a reference to the CSA battle flag.

Put 1 ball in each jars first, total used 4 balls.

Then there are 56 ways to put the remaining n = N-4 balls in the 4 jars.

Using the stars and bars method, putting 3 bars in n+3 slots, to distribute the remaining n balls into 4 jars.

${n+3 \choose 3}=56 \rightarrow (n+1)(n+2)(n+3) = 336 \rightarrow n+1=6$

$N = n + 4 = \boxed{9}$

This sequence appears in Pascal's triangle