Permutation and combination

Question

In how many ways $11$ identical toys be placed in $3$ distinct boxes such that no box is empty?

My answer is stupid:

Why I should not put ${}_{11}C_3$. Why should I solve this with ${}_{10}C_2$ as in my book. Can anyone suggest me some good material for permutation and combination on net!

Answer 1

The toys are all identical. So, all you are trying to do is to choose the NUMBER of toys that get put in each box.

The condition given says that you must put at least one toy in each box; so, you have 8 identical toys remaining, and can distribute them between the three boxes in any way you like.

This can be thought of using a 'stars and bars' argument: any arrangement of these 8 toys is equivalent to a sequence of 8 $*$'s and 2 $|$'s, where each $*$ represents a toy and the $|$'s represent the cutoffs between boxes. For instance, $$ **|*****|* $$ represents the first box getting two of these eight toys, the second getting five, and the third getting one.

The number of such sequences is $\binom{10}{2}$: there are 10 positions total, and we need to choose which 2 of those positions are bars.

Answer 2

When you solve it with 11C3 there involves a case where a box is left empty and also a case where two box are left empty while in case of 10C2 means that one ball is already placed in box and we are looking for places of other 10 balls for 2 boxes.