Question

Three boys picked five apples. In how many ways can these five apples be distributed among the three boys so that each can have any number, of course, not exceeding five? (All apples are considered the same.)

Answer 1

Let us consider, in addition to 5 apples, 2 new things say oranges (this number is one less than the number of boys). Then the number of ways of distribution will correspond to the number of arrangements of five A’s and two O’s.

For example, consider the following arrangements:

(a) I OAAA l O AA; draw line just before O’s to partition this arrangement. This arrangement corresponds to 0 + 3 + 2, (the numbers corresponding to the number of A’s before the line of demarcation, between consecutive lines and after the last line). 

(b)  I O A I O AAAA corresponds to 0 + 1 + 4

(c)  A I OA I O AAA corresponds to 1 + 1 + 3

The number of permutation = 7!/(5!2!)  or ⁷C₂

Alternative Solution: If n apples are distributed among r boys, the number of ways of distribution would be ^{n + (r – 1)}Cr _{– 1}. Hence, in this case number of ways are  ⁷C₂