sorry I'm not sure if the title of my question would best describe this problem. Well, I tried to solve the problem below, but not sure with my answer.
Here's the question,
In how many ways can nine different pieces of candy be given to three children so that one child gets two pieces, another child gets three pieces, and another gets four pieces?
Solution:
Distributing 2 pieces of candy to child 1 is same as distributing 2 pieces of candy to child 2 and child 3, since all of them are identical.
If they were distinct, then we can simply solve it by${{9}\choose{2,3,4}}.$Since 2 pieces of candies can be on either of the 3 children. Thus, there are 3 different children for 2 pieces of candy to distribute on, and there are (3-1) children for 3 pieces of candy to be distributed, and there are (3-2) children for the 4 pieces of candy to be distributed. If this is the case, then there are${{9}\choose{2,3,4}}\cdot 3!\qquad \blacksquare$I'm not sure with my solution, and I want to ask for your idea guys. Thanks in advance!