Here, it is said that there is at least one ball in each box, thus, there can be only two combinations possible – (3.1.1) or (1,2,2).

for the first combination (3.1.1), there are $\binom{5}{3}$ ways to chose 3 balls, then 3 ways to put these 3 balls into one box. for the remaining balls, there are 2 ways for the 4^{th} ball and 1 way for the last ball. Applying product rule, we get $\binom{5}{3}$ *3 *2 = 60 ways

for the second combination (1,2,2), there are 3 ways to choose the box containing the lonely ball and 5 ways to choose which ball will be the unlucky one. For the remaining balls, there are $\binom{4}{2} $* $\binom{2}{2} $ ways. Applying product rule we get 3*5*$\binom{4}{2} $ * $\binom{2}{2} $ = 90 ways.

Applying sum rule(because there are two alternatives), we get 60+90 = 150 ways.

This is similar to the Integer composition problem. We need to find the number of compositions of 8 into 3 parts.

The first is (1.1.3) which can be done in $\binom{5}{3}$ = 10 ways.

for the second composition, that is (1,2,2), we csn do it in ($\binom{5}{2} $ * $\binom{4}{2} $) / 2 = 15 ways.

Applying Sum rule, we get, 10+15 = 25 ways.

This is a Stars and Bars Probelm, a very common and very important IODB template. There are 3-1=2 bars and 2 stars. (there is one ball each in the boxes already, thus 5-3 = 2). To put the 2 balls, there are 4C2 or 6 ways.

This is the Integer Partition Problem. We need to find the number of partions of 5 into 3 parts. And as we have already seen, there 2 partitions. 1+1+3 and 1+2+2. Thus, there are only 2 ways.