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There are $ 3 $ boys and $ 7 $ girls. How many ways can we divide them into three unlabeled groups such that each group has a boy, two groups have three people, and the third group has four people?

I think that the answer is $\frac{3!\times \binom{7}{2} \times \binom{5}{2}}{2}=630$, but other people said that it is $\frac{3\times \binom{7}{2} \times \binom{5}{2}}{3}=210$.

Edit: I did wrote something incorrect.

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  • $\begingroup$ Why are you dividing by $2$ exactly? $\endgroup$ – Dietrich Burde Jan 1 '20 at 15:11
  • $\begingroup$ Are the groups labeled? Specifically: if we switch the two smaller groups, do we get a new arrangement or not? $\endgroup$ – lulu Jan 1 '20 at 15:15
  • $\begingroup$ ${7\choose2}\times{5\choose2}=210$,, so your arithmetic is wrong in both cases. $\endgroup$ – almagest Jan 1 '20 at 15:15
  • $\begingroup$ @lulu Labelled? These are groups of individuals - surely we can assume they are distinguishable. There is no suggestion that the groups are ordered. $\endgroup$ – almagest Jan 1 '20 at 15:16
  • $\begingroup$ @almagest Ordered then. As I specified, my question was "are the two arrangements $(b_1,g_1,g_2), (b_2,g_3,g_4), (b_3, g_5, g_6, g_7)$ and $(b_2,g_3,g_4),(b_1,g_1,g_2), (b_3, g_5, g_6, g_7)$ different or not?" As others have asked about division by $2$, I think it's fair to say that the point is not obvious. $\endgroup$ – lulu Jan 1 '20 at 15:19
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You're just identifying which of the girls go along with each boy.

So there are 3 ways to choose which of the boys is in the bigger group, and then $\binom73$ ways to choose whose his partners are. Then there are $\binom42$ ways to choose which of the remaining girls goes with the older of the remaining boys. Thus, the total is $3\cdot\binom73\binom42=3\cdot35\cdot6=630$.


The reason we don't need to do any dividing is because every grouping we are making are distinct. If Adam, Ben, and Carl are the three boys, then putting Ben with Jane and Kim and Carl with Laura and Mary is a different grouping than if put Ben with Laura and Mary and Carl with Jane and Kim. Each of the three groups CAN be distinguished -- by which boy is in the group -- so there is no need to worry about overcounting.

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All $3$ boys have to be in different groups. Then we have $3$ choices which boy is in a group of $4$ and $\binom7{3,2,2}=210$ ways to choose girls for the groups (assuming the groups are unlabelled), for a total of $3\cdot210=630$.

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Since groups are unlabeled 3 boys can be distributed only in 1 way and as 7 girls to be distributed as 2,2,3 into groups hence only 2 of the group be treated like identical groups. = 630 For more similar to this there is question in other forum link is https://www.mathsdiscussion.com/forum/topic/permutations-and-combinations-2/?part=1#postid-73

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  • $\begingroup$ Your answer is incorrect since it matters which boy is in the group of four people. $\endgroup$ – N. F. Taussig Jan 1 '20 at 16:47
  • $\begingroup$ What is Number of ways 52 playing cards be divided into 4 parts of 15,15,15 and 7 . $\endgroup$ – mathsdiscussion.com Jan 1 '20 at 17:04
  • $\begingroup$ That is not an analogous problem since the groups are distinguished by which boy is in the group with four people. $\endgroup$ – N. F. Taussig Jan 1 '20 at 17:05
  • $\begingroup$ In a question it is mentioned that groups are unmarked. Hence 3 boy can be distributed into 3 unmarked groups in 1 way now groups becomes distinct and of which two will be like identical as number of girls will be equal in two of the group. $\endgroup$ – mathsdiscussion.com Jan 1 '20 at 17:10
  • $\begingroup$ The group of four is distinguished by its size. $\endgroup$ – N. F. Taussig Jan 1 '20 at 17:12
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Choose 1 boy from 3, the 2 girls from 7 and 2 girls from 5 and then 3 girls from the 3 left.

So, total combinations: $ ^3C_1 \times ^7C_2 \times ^5C_2 =630$

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