8,9

Six boxes are numbered from 1 to 6. In how many ways can 20 identical balls be distributed into these boxes

a) so that no box is left empty?

b) if some boxes may be left empty)?
a) Let's lay out the balls in a row. To determine the arrangement of our balls in six boxes, we will divide the row with five partitions into six groups: the first group for the first box, the second for the second box, and so on. Thus, the number of ways to arrange the balls in the boxes is equal to the number of ways to place five partitions. The partitions can stand in any of the 19 positions (between 20 balls - 19 gaps). Therefore, the number of their possible placements is $\mathrm{C}_{19}^{5}$.

b) Consider a row of 25 items: 20 balls and 5 partitions, arranged in any order. Each such row uniquely corresponds to a certain way of arranging the balls in the boxes: the first box contains the balls located to the left of the first partition, the second box contains the balls located between the first and second partitions, and so on (there may be no balls between some partitions). Therefore, the number of ways to arrange the balls in the boxes is equal to the number of different rows of 20 balls and 5 partitions, which is $\mathrm{C}_{25}^{5}$.

## Answer

a) $\mathrm{C}_{19}^{5}$;
b) $\mathrm{C}_{25}^{5}$ ways.

In how many ways can a natural number $n$ be represented as a sum of
a) $k$ natural summands?

b) $k$ non-negative integer summands?

(Representations differing in the order of the summands are considered different.)

## Hint

Represent $n$ as a sum of $n$ ones: $n=1+1+\ldots+1$. Now let's call these $n$ ones "balls," and the $k$ summands from the problem statement - "boxes." See problem 30717 for further details.

## Answer

a) $\mathrm{C}_{n-1}^{k-1}$;
b) $\mathrm{C}_{n+k-1}^{n}$ ways.