Apples arrangement

One way to tackle this problem is by invoking the binomial theorem. Too long and unnecessary so we switch to Method 2 - the good old case consideration method.

The constraints are that every boy gets at least 1 apple and gets no more than 4.

Cases thus are :

1. ( 4,3,1). This distribution sequence can be arranged in 3! = 6 ways. For each of the above distributions, the apples can be chosen ( since apples are distinct!) in $\binom{8}{4}$ * $\binom{4}{3}$ * $\binom{1}{1}$ ways.

2. (3,3,2). This distribution sequence can be arranged in $\large \frac{3!}{2!}$ = 3 ways. The apples can be chosen in $\binom{8}{3}$ * $\binom{5}{3}$ * $\binom{2}{2}$ ways.

3 .(2,2,4). This distribution sequence can also be arranged in $\large \frac{3!}{2!}$ = 3 ways. We choose apples in $\binom{8}{2}$ * $\binom{6}{2}$ * $\binom{4}{4}$ ways.

Evaluate the total cases hence.