How many ways can divide 12 different books among 4 children, so that
- each child reaches 3 books?
- each of two older children get 4 books and each of two younger children get 2 books?
I think the answer is:
1)
We can think of this as arranging the 12 books in a row and then dividing them into groups of 3, where the order within each group does not matter. We can use the formula for combinations to find the number of ways to do this:
C(12,3) x C(9,3) x C(6,3) x C(3,3) = 739,200.
we can approach this problem by dividing it into two steps. First, we choose the 4 books for each of the two older children, and then we choose 2 books for each of the two younger children.
To choose the 4 books for each of the two older children (order in which we choose the books does not matter):
C(12,4) x C(8,4) = 34,650
We have 4 books left. For the first younger child, we have 4 books to choose from, so we have C(4,2) = 6. For the second younger child, we have 2 books left, so we have C(2,2) = 1. Finally, We get 34,650 x 6
I'm not sure my answer is right.
I think it's a little easier than you're making it. In each case, line up the books from left to right and always assign the books in the same positions to the appropriate kid. In other words, in case $1$, the oldest kid gets the first three books, the next oldest kid gets the next three books, and so on.
There are $12!$ ways to arrange the books in a line. In the first case, the order within each group of three is irrelevant so your answer is $\dfrac{12!}{(3!)^4}=369600$. (I believe there's an arithmetic error in your first calculation.) For similar reasons, the answer to case $2$ is $\dfrac{12!}{(4!)^2(2!)^2}=207900$.