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Four boys and three girls want to sit on a bench. In how many ways can they sit:

  • if the boys are next to one another and the girls are next to one another.
    Since boys and girls are separated, they are treated as two units. The first unit—boys—consist of 4 people, and the order matters, so $_4P_4$ is 24. The second unit—girls—consist of 3 people, and the order matters, so $_3P_3$ is 6. Then, as they are two units, we multiply them together, giving us an answer of 144 ways.
  • if there are neither boys next to one another nor girls next to one another.
    There are four possible seats for boys, each has the possibility of hosting four different boys. Then, there are $4^{4}$=256 ways for them to sit. There are three possible seats for girls, each has the possibility of hosting three different girls. Then, there are $3^{3}$=27 ways for them to sit. If we merge them altogether, we get 256*27=6912 ways overall.

I am pretty confident about the first one, but I have lots of doubts about the second one. I, initially, thought: Isn't the second question similar to the first one? Then, I brushed it off as I contemplated over it for a few seconds.

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    $\begingroup$ For the first one, maybe the boys sit on the left, maybe they sit on the right. For the second, there are only $7!=5040$ ways to sit the children with no constraints, so your answer is too large. $\endgroup$
    – lulu
    Commented May 28, 2023 at 13:56
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    $\begingroup$ For the second, the only possible pattern is $BGBGBGB$ so you just need to populate that. $\endgroup$
    – lulu
    Commented May 28, 2023 at 13:58
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    $\begingroup$ I don't understand your comment on the second one. There is only the one pattern, so the answer is just $4!\times 3!$ $\endgroup$
    – lulu
    Commented May 28, 2023 at 14:06
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    $\begingroup$ My point with the $7!$ was that your answer was obviously too big. $7!$ is the number of ways to seat the kids with no rules at all. If you have rules, the number must get smaller. In the first scenario, you have two patterns, namely $BBBBGGG$ and $GGGBBBB$. In the second you have only one, namely $BGBGBGB$. Thus there are twice as many ways to handle the first scenario as there are ways to handle the second. $\endgroup$
    – lulu
    Commented May 28, 2023 at 15:29
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    $\begingroup$ To stress: Any fixed pattern can be populated in exactly $4!\times 3!$ ways. You just assign particular boys to the boy seats and particular girls to the girl seats. Thus, if you wanted $BBGGGBB$ there would be $4!\times 3!$ ways to get it. so the point really is to count the "good" patterns. That is, those that satisfy whatever rules you have in mind. $\endgroup$
    – lulu
    Commented May 28, 2023 at 15:31

1 Answer 1

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If there are neither boys next to one another nor girls next to one another, you are looking for the number of ways of arranging the letters B (representing boys) and G (representing girls) in a 7-letter word in such a way that there are no two equal letters next to each other, the letters are distinguishable. The only possiblity is to start with a B since there are less girls than boys. Positions 1,3,5 and 7 will be occupied by boys. And positions 2,4 and 6 by girls. For the first seat, you have 4 different options; for the third seat 3 different options; for the fifth seat only 2 different options and finally for the seventh seat the final option (1 option). Similarly for the girls, for the second seat you have 3 different option; for the forth only 2 options and the sixth seat will be determined by the previous 2, hence only one option. The total number of possible ways of satifying the condition is: $4\times3\times3\times2\times2\times1\times1=144$

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