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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!

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I think you have it a bit backwards. $\binom{9}{2,3,4}$ does not distinguish between the subsets but $\binom{9}{2,3,4}\cdot 3!$ does. –  EuYu Sep 20 '12 at 15:06
    
I think so, but ${{9}\choose{2,3,4}}$ was done using the multinomial theorem, which takes the coefficient of the $x^2y^3z^4$. And if these x,y, and z are identical, then there would be repetitions. –  Al-Ahmadgaid Asaad Sep 20 '12 at 15:13
    
The answer is OK, probably your reasoning was OK. The explanation of the reasoning is unclear. By the way, there is a default assumption in such problems that people (and even children) are distinct. –  André Nicolas Sep 20 '12 at 15:14
    
Yeah true Andre, it's unclear, well my aim is just to arrive on my final answer. I can't consider that default assumption, since the first problem before this one was almost the same, the only thing is that each child has a name (Kim, Ron, and Bob). Thus, if I consider that, then I would have the same answer with the first one I mentioned. –  Al-Ahmadgaid Asaad Sep 20 '12 at 15:22

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