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Ten girls are to be divided into groups of sizes $3,3,2,2$. Also, there are $3$ boys. Number of ways of linear sitting arrangement such that between any two groups of girls, there is exactly one boy (no boy sits at either extreme end)?

MY SOLUTION:

$10$ girls can be divided into groups of sizes $3,3,2,2$ in $$\frac{10!}{3!3!2!2!2!2!}$$ ways which gives me unique combination of groups.

I can then arrange these groups in $4!$ ways, and people within them in $3!3!2!2!$ ways.

Finally the $3$ boys in $3!$ ways.

Seems correct? It gives $$\frac{10! 4! 3!}{2! 2!}$$ ways.

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  • $\begingroup$ Yes your approach seems absolutely fine! Its good to show your work when you ask a question. $\endgroup$ – samjoe Oct 2 '17 at 7:31
  • $\begingroup$ @samjoe but I am skeptical about arranging groups. Should it be multiplied by 4!? I am doing so because each group so formed is unique in its configuration. It shouldn't lead to duplicate arrangements. Will it? $\endgroup$ – Ajax Oct 2 '17 at 7:39
  • $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Oct 2 '17 at 9:10
  • $\begingroup$ @N.F.Taussig Alright, Thanks! Do you think my approach is correct for this question? And that I am not counting duplicate cases? $\endgroup$ – Ajax Oct 2 '17 at 9:28
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Assuming the girls have sat down, they leave 3 gaps between them. 1 for each boy. Thus the first boy can pick between 3 chairs, the second boy 2 chairs, and the third doesn't get to pick. So there are $3\cdot 2=3!=6$ ways the boys can sit. Now the girls are a little bit more tricky. Notice that it isn't specified how the girls are to be divided among the groups, thus the first girl can pick among 10 spots, the next 9 and so on. Finally we have to account for the ways the 4 groups can be arranged, which by the binomialcoefficient is equal to$\frac{4!}{2!2!}$.

Hence your final answer is $$ 3!\cdot 10!\cdot \frac{4!}{2!2!} $$

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Since there are $4$ groups of girls, and three boys, there is only one case possible for boys to sit between the groups.

Boys can be arranged in $3!$ ways in their seats, the groups of girls can be arranged in $\frac{4!}{2! 2!}$ ways. For any such arrangement, girls can be rearranged in $10!$ ways.

So the answer should be: $$3!\cdot\frac{4!}{2!2!} \cdot10!$$

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For $10$ girls we have $10!$ permutations. We have $3!$ for boys. We just put the boys in the right positions. Thus, the result is $10! \times 3!$.

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  • $\begingroup$ Hasan Groups can also be distributed around boys. I mean the size of groups is not identical, that is.. $\endgroup$ – samjoe Oct 2 '17 at 7:11
  • $\begingroup$ So, this answer 10!x3! is for any combination of groups (totalling to 10)? $\endgroup$ – Ajax Oct 2 '17 at 7:13
  • $\begingroup$ @Ajax, What is the meaning of 'totaling to 10'? $\endgroup$ – Hasan Heydari Oct 2 '17 at 7:20
  • $\begingroup$ 1,2,2,5 or 2,2,2,4 or 3,3,3,1 all total to 10 $\endgroup$ – Ajax Oct 2 '17 at 7:28
  • $\begingroup$ Your answer is incorrect because you have not accounted for the number of ways of arranging two groups of three girls and two groups of two girls. $\endgroup$ – N. F. Taussig Oct 2 '17 at 8:56
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Your solution is correct.

Here is another way to think about the problem:

We will seat the girls in two blocks of three seats and two blocks of two seats. There are $\binom{4}{2}$ ways to choose the positions of the blocks of three seats. Once the blocks of seats have been arranged, there are $\binom{10}{3}$ ways to select the girls who will sit in the leftmost block of three seats and $3!$ ways to arrange them in those seats, $\binom{7}{3}$ ways to select which of the remaining seven girls who will sit in the other block of three seats and $3!$ ways to arrange them in those seats, $\binom{4}{2}$ to select which of the four remaining girls will sit in the leftmost block of two seats and $2!$ ways to arrange them in those seats, and $2!$ ways to arrange the girls who will sit in the other block of two seats. The three boys can be arranged in the three seats that separate the blocks of girls in $3!$ ways. $$\binom{4}{2}\binom{10}{3} \binom{7}{3}\binom{4}{2}\binom{2}{2}3!3!2!2!3! = \frac{4!}{2!2!} \cdot \frac{10!}{3!7!} \cdot \frac{7!}{3!4!} \cdot \frac{4!}{2!2!} \cdot \frac{2!}{2!0!} \cdot 3!3!2!2!3! = \frac{4!}{2!2!} \cdot 10! \cdot 3!$$ as you found.

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