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If four boys and four girls play tricks, how many ways can they join hands, provided that at least two girls are together?

My plan is to determine the circular permutation of the eight (boys + girls), which is equal to 7! and to subtract from it the cases in which at least two girls are separated. That is, only two separate girls, only three separate girls or only four separate girls. How to proceed with this approach?

The answer is 7! - 3! * 4! which, by the presence of less, suggests the use of the complementary combination. To get to 3! * 4 !, I believe that it is enough to calculate the circular permutation of the boys (which will be among the girls), which results in (4 - 1)! = 3 !, and multiply by the girls' permutation: 4 !. Why, considering that the boys should be among the girls, it is not correct to multiply the circular permutations of the two, getting 3! * 3! ?

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  • $\begingroup$ If you want to find the number of arrangements with at least two girls together, you should be subtracting the number of arrangements with no two girls together. $\endgroup$ May 23, 2019 at 17:18
  • $\begingroup$ Yes, I know. I just wrote wrong, thank you very much for the correction. $\endgroup$
    – gmn_1450
    May 23, 2019 at 17:21

2 Answers 2

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For these circular permutation questions, you can avoid "division by symmetry" arguments by finding some sort of focal point for your circle and placing everyone else around the focal point. This can be accomplished by taking "the youngest" or the "smallest" or some other arbitrary method of selection and placing that person or object into the circle in an arbitrary position first.

So, for this, let us take the youngest boy and have him "start" the circle. Now that he is there, we can treat the remaining positions the same as a line. The only way to have it be so that no two girls in this scenario are for it to repeat "boy-girl-boy-girl-boy-girl-..." so let us decide who the girl standing to the right of our youngest boy is. This girl can be chosen in $4$ ways. Then, pick the boy who stands to her right. Then pick the girl to his right, and so on so forth until we are done.

This gives $4\times 3\times 3\times 2\times 2\times 1\times 1 = 4!\times 3!$, no division necessary.

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The boys and girls are in the same circle. If you used $3!$ twice, you would be dividing by $4$ twice, once for the boys and once for the girls, which is only valid if they are each in their own independent circles. Since they are only in one circle, there is only one degree of freedom, so you should only divide by $4$ once to eliminate over-counting; this results in $(4!\times 4!)/4=4!\times 3!$.


Look at it this way; if you were instead counting linear arrangements of boys and girls where no two girls are together, and also the ends are not both girls, there would only be two possible patterns: $$ BGBGBGBG\qquad \text{ and }\qquad GBGBGBGB $$ Each of these can be completed in $4!\times 4!$ ways, so there are $2\times 4!\times 4!$ linear arrangements. However, since you want circular arrangements, you need to divide by $8$, since the linear arrangements are partitioned into groups of size $8$ which all represent the same circular arrangement. Therefore, the answer is $(2\times 4!\times 4!)/8=3!\times 4!$.

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