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We would like to count how many ways 3 boys and 3 girls can sit in a row. How many ways can this be done if:

(b) all the girls sit together? ...

(c) every boy sits next to at least one other girl?

ANSWER: If three boys sit next to each other, no combinations work. If two boys sit next to each other, it fails if and only if the pair of boys sitting next to each other are on an edge (ie. BBGGBG, BGGGBB). If no two boys sit next to each other, all combinations work. There are 4!3! combinations with three boys together (see part (b)). If we place two boys on the edge, we have two choices, left or right to place them. We then choose the position of the third boy from threeremaining positions (he can't be next to the two other boys) for a total of 2*3*3!3! positions (3!3! to account for varying positions of unique boys and girls). Since there are 6! total positionings, there are 6!- 4!3!- 2*3*3!3! = 360 positionings where no two boys sit next to each other.

i get where 6!- 4!3! comes from but don't understand where 2*3*3!3! comes from

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Of course, if the kids are too shy, opposite-sex people never get to sit next to each other. –  Ahaan Rungta Feb 14 at 17:57
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1 Answer

I think it is explained in your OP. We do the same thing with a bit more detail.

Since we have already counted the number of "bad" positions with all the boys together, it remains to count the number of bad positions in which the boys are not all together, but some boy is not next to a girl.

There must be two boys together, and they must be at the left end or the right end ($2$ choices). Take such a choice, say left end. Then the remaining boy can be in any one of $3$ places, namely positions $4$, $5$, or $6$. That gives us $(2)(3)$ ways of choosing the seats the boys will occupy. For each of these $(2)(3)$ choices, there are $3!$ ways of permuting the boys among the chosen seats, and for every such choice there are $3!$ ways of permuting the girls, for a total of $(2)(3)(3!)(3!)$.

Another way: Counting all possible arrangements and subtracting the "bad" ones is often good strategy. But let us count directly.

Our condition will be satisfied if either no two boys are together, or exactly two boys are together and they are not in the end seats.

Count first the arrangements in which no two boys are adjacent. Write down $G\quad G\quad G$. This determines $4$ "gaps" into which we can slip the boys, one boy per gap. There are $4$ "gaps" because we are including the end gaps. There are $\binom{4}{3}$ ways of choosing $3$ of these gaps.

Or else we could slip $2$ boys into one of the two center gaps ($2$ choices), and then slip the remaining boy into one of the $3$ remaining gaps, for a total of $6$ choices.

Thus the places for the boys can be chosen in $10$ ways. For each of these ways, we can arrange the boys is $3!$ ways, and then the girls in $3!$ ways, for a total of $360$.

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