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How many ways can 10 people be split into groups of 2 and 3? the answer says ${5 \choose 2}$... But isn't it the answer if the question were: "How many ways can 5 people be split into groups of 2 and 3?"

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We can have the groups assigned with the following number of people: $$(1)\,\{2,2,2,2,2\}\\(2)\,\{2,2,3,3\}$$

Case 1: Choose groups of $2$ from $10, 8, 6, 4$ in succession until we reach our final pair. Our $5$ sets of groups are the same irrespective of the order in which they are chosen, so the number of ways here would be: $$\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\frac{1}{5!}$$ Case 2:

  1. Choose $3$ people from $10$.
  2. Choose $3$ people from the remaining $7$.
  3. Choose $2$ people from the remaining $4$, leaving the remainder to form the final pair.

In summary, we have $2$ different types of groups, each containing $2$ groups. Eliminating repeats by accounting for the order in which such groups can be chosen, the number of ways here is: $$\binom{10}{3}\binom{7}{3}\binom{4}{2}\left(\frac{4!}{2!\,2!}\right)^{-1}$$

Add the cases together and you have your result.

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    $\begingroup$ The factor $\left(\frac{4!}{2!\,2!}\right)^{−1}=\frac{1}{6}$ is not right. There are four repeats of each partition, not six. For example consider selecting from the set $\{A,B,C,D,E,F,G,H,I,J\}.$ The selection $$(\{A,B,C\},\,\{D,E,F\},\,\{G,H\},\,\{I,J\})$$ gives the same partition as the selection $$(\{A,B,C\},\,\{D,E,F\},\,\{I,J\},\,\{G,H\}),$$ the selection $$(\{D,E,F\},\,\{A,B,C\},\,\{G,H\},\,\{I,J\}),$$ and the selection $$(\{D,E,F\},\,\{A,B,C\},\,\{I,J\},\,\{G,H\}).$$ $\endgroup$ Dec 12, 2014 at 19:02
  • $\begingroup$ The multinomial coefficient $$\binom{10}{3}\binom{7}{3}\binom{4}{2}=\binom{10}{3,\,3,\,2,\,2}$$ is not counting the ways of choosing two groups of $3$ and two groups of $2$ in any order; it's counting the ways of choosing a group of $3,$ than another group of $3,$ then a group of $2,$ then another group of $2,$ in that order. The overcounting results from the possibility that one selection may have the same groups of $3$ as another, but in the opposite order, and likewise for the groups of $2.$ $\endgroup$ Dec 12, 2014 at 19:02
  • $\begingroup$ I gave a slight variation of your answer. Hope you do not mind. $\endgroup$
    – mick
    Feb 21, 2018 at 15:57
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We can have the groups assigned with the following number of people: $$(1)\,\{2,2,2,2,2\}\\(2)\,\{2,2,3,3\}$$

Case 1: Choose groups of $2$ from $10, 8, 6, 4$ in succession until we reach our final pair. Our $5$ sets of groups are the same irrespective of the order in which they are chosen, so the number of ways here would be: $$\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\frac{1}{5!} = 945$$

Notice we can do this in another way.

$10/2 = 5$ Therefore there are $5$ groups. We can arrange 10 people in 10! Ways. If we pick the $n,n+1$ persons of that arrangement Then we picked 5 times groeps of size $2$. The order of the 5 groups needs to be removed So we correct $10!$ to $\frac{10!}{5!}$. The order of the 2 members in every group needs to be removed too , So we correct $\frac{10!}{5!}$ to

$$ \frac{10!}{5! 2^5} = 945 $$.

This method gives the same result ($945$).

Case 2 can be done in a similar way.

This answer is similar to the accepted one , but I Just wanted to show an alternative way.

Since this alternative way is also an answer and this whole is too long for a comment I made it into an answer.

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This is equivalent to splitting a group of 10 people into a group of 5, a group of 3, and a group of 2. There are $10!/(5!3!2!)$ ways to do this. Equivalently, you could think of first choosing five people out of the $10$, and then choosing two people out of the $5$. There are $\binom{10}{5}\cdot\binom{5}{2}$ ways to do this. If you compute both methods give you $2520$.

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  • $\begingroup$ thank you! Is this mean the provided answer (by me )was wrong? $\endgroup$
    – user188750
    Dec 1, 2014 at 7:55
  • $\begingroup$ Aha perhaps I interpreted the question incorrectly - this solution is for choosing a group of $2$ and a group of $3$ from the $10$, whereas perhaps you are asking how to split the entire group of $10$ into pairs and triplets as the other answer indicates. $\endgroup$
    – ajd
    Dec 1, 2014 at 7:58

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