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We have $n$ identical objects, and we want to distribute them to $3$ different children A, B, C. In how many ways can this assignment be done?

Please note that for example if we had $6$ objects, if A gets $4$, B gets $2$ and C gets $0$, it's different from A gets $4$, B gets $0$ and C gets $2$. Compute the general formula and then find the correct answer if $n=9$.

Answer: 84

No matter what, I still don't get the correct answer. I tried as such: I lined up the $n$ objects , (let's draw them as $0$s). Now I want to separate them into $3$ teams. I am going to throw randomly two sticks ( imagine them as $1$s) . Where they fall, they will form the limits of each child's objects. For example $01 0 0 01 0 0 0 $ : Now take a look at left if the first stick: That's going to be how many objects A gets ( here is 1), between the first stick and the second it's what B gets, and what remains goes to C. If we had : $1 0 0 0 0 0 0 0 1$ A gets $0$, B gets everything, C gets $0$, if we had $ 110 0 0 0 0 0 0$ ** A gets $0$, B gets $0$, C gets it all** and so on... So, we actually need to compute how to throw 2 identical sticks into $(n-1) + 2 = (n+1)$ holes, which can be done in $C(n+1 + 2 -1, 2)=C(n+2,2)$ which for $n=9 \rightarrow C(11,2)=55$. What am I missing?

Khashayar Baghizadeh
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tonythestark
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    You may find this useful: https://en.m.wikipedia.org/wiki/Composition_(combinatorics) – cgss Aug 30 '20 at 16:40
  • Consider this way . Choose a random number as n such as 7 then take A, B, C assignment first 0 to A then 0 to B then 7 to C. Start filling up all types of ways in which you can assign them. Slowly a pattern will start emerging and then try to understand it also then try for more numbers and atlast go for n. Hope it help. Please upvote. – SAGNIK UPADHYAY Aug 30 '20 at 16:41
  • This may help you. – SarGe Aug 30 '20 at 16:48
  • @SarGe This is exactly what OP already did, why are you pointing OP in this direction? – Christoph Aug 30 '20 at 17:53
  • @SAGNIKUPADHYAY Please do not beg for upvotes on this site and note that comment votes don't contribute to reputation anyway. – Christoph Aug 30 '20 at 17:54
  • @Christoph Two sequences that differ in the order of their terms define different compositions of their sum, while they are considered to define the same partition of that number., from Wiki. – cgss Aug 30 '20 at 18:11
  • @cgss As a (strict) composition $9+0+0$ doesn't even come up, since $0$ is not allowed as a summand. This would be a weak composition. But you are right, they are also mentioned in the article. – Christoph Aug 30 '20 at 19:15
  • Yes, exactly. I didn't mean that the answer was a composition. I meant for OP to read it and find the weak composition. – cgss Aug 30 '20 at 20:11

1 Answers1

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The number of ways how to distribute $n$ identical objects among $k$ persons is identical to the number of $k$-tuples of non-negative integers that sum to $n$. By stars and bars this number is given by $$ \binom{n+k-1}{k-1} = \binom{9+3-1}{3-1} = \binom{11}{2} = 55. $$ Your answer is correct, the given answer is wrong. (It coincides with $\binom{9}{3}$ by the way.)

Christoph
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