Distributing different toys

Question

Find the number of ways in which 12 different toys may be distributed among 4 children so that each child gets at least 2 toy.

Answer 1

Count the ways to distribute the 12 toys between 4 kids for each possible grouping:

IE: (1 kid gets 6 toys, 3 get 2), (1 gets 5, 1 gets 3, 2 get 2), (2 get 4, 2 get 2), (1 gets 4, 2 get 3, 1 gets 1), (4 get 3).

[edit: to clarify]Select kids to receive each amount of toys, then count the ways to arrange the toys among these groupings by permutating all 12 and dividing by the permutation of each group (since the order each kid obtains her toys does not matter).

1. $N(6,2,2,2) = \frac{12!}{6!2!2!2!}{4\choose 1} = 83160 $
2. $N(5,3,2,2)=\frac{12!}{5!3!2!2!}{4\choose 1}{3\choose 1} = 1995840$
3. $N(4,4,2,2)=\frac{12!}{4!4!2!2!}{4\choose 2}=1247400$
4. $N(4,3,3,2)=\frac{12!}{4!3!3!2!}{4\choose 1}{3\choose 2}=9979200$
5. $N(3,3,3,3)=\frac{12!}{3!3!3!3!}=369600$

Total: $$ = 13,675,200$$

Answer 2

Alternatively:

There are $4^{12}$ ways to assign each toy to a kid.

We need to exclude cases where a kid receives 1 or 0 toys. $ 4 \left({12 \choose 1} 3^{11} +3^{12}\right)$

We need to include cases where this happens to 2 kids: ${4\choose 2}\left({12\choose 1}{11\choose 1} 2^{10} + 2 {12\choose 1} 2^{11} + 2^{12}\right)$

We need to exclude cases where this happens to 3 kids: ${4\choose 3}\left({12\choose 1}{11\choose 1}{10\choose 1} 1^{9} + {3 \choose 2}{12\choose 1}{11\choose 1} 1^{10} + {3\choose 1}{12\choose 1} 1^{11}+ 1^{12} \right)$

So by inclusion-exclusion: $$4^{12}-4\left(\frac{12!3^{11}}{1!11!}+3^{12}\right)+\frac{4!}{2!2!}\left(\frac{12!2^{10}}{10!}+2\frac{12!2^{11}}{11!}+2^{12}\right)-\frac{4!}{3!}\left(\frac{12!}{9!}+\frac{3!12!}{2!10!}+\frac{3!12!}{2!11!}+1\right)$$

$$=4^{12}-4\left(12\times 3^{11}+3^{12}\right)+6\left(12\times 11\times 2^{10}+12\times 2^{12}+2^{12}\right)-4\left(12\times 11\times 10+3\times 12\times 11+3\times 12+1\right)$$

$$=7271880$$