I am not the best at math (very bad at probability) and explaning things, but I'll give this a shot.

The book method seems to be correct. They overcounted and then subtracted away the impossible situations dictated by the requirements.

For your method, there is one slight problem with it. Let us consider the first boy, call him Bob. Assume that Bob receives only two fruits (unlucky that he is), and that he has received F_{1} and F_{2}.

Your solution takes into account two distributions. The first is only to ensure that each boy receives at least one fruit, and the second to distribute the rest of the fruit. Let's call the set of fruit Bob receives from the first distribution S and the set of fruit from the second distribution T.

By multiplying the two distributions together, you have counted

S = {F_{1}}, T = {F_{2}} and

S = {F_{2}}, T = {F_{1}}

as two different situations when they should only be regarded as one.

This may not seem much, but if Bob received three fruits, there would be an overcounting of 3 for each possible way Bob can receive three fruits, i.e. _{12}*C*_{3}. For four fruits, there is an overcounting of 3 for each possible way Bob can receive four fruits.

In addition, each time there was a miscount for Bob, there is also a huge miscount for Fred and George also. Furthermore, after all that, we must also count the miscounts for Fred and George.

Therefore, the miscount is quite large. I believe that is the problem in your solution.