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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? Since all the girls must sit together, we treat the girls as a single unit. Then we have 4 people to arrange with 3! positions for 3 girls for a total of 4!3! ways to arrange them.

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You’re forgetting that the girls can sit in $3!$ different orders within their block. –  Brian M. Scott Jan 27 '13 at 0:53
    
The $4$ is what gave rise to your $4!$. –  André Nicolas Jan 27 '13 at 0:56
    
thanks brian, i got it. 3! is the different ways the girls can sit next to each other since they are all different individuals –  user59795 Jan 27 '13 at 0:57
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3 Answers 3

There are $4!$ ways to arrange the four blocks, where each boy is one block, and the three girls together are the fourth block. Once you’ve done that, you can arrange the $3$ girls in $3!$ different orders within their block. Thus, for each arrangement of the blocks you get $3!$ arrangements of the people, for a total of $4!3!$ arrangements of the people.

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Another way is to observe, as you did, that there are $4$ legal arrangements of the letters $b$ and $g$. For each of these arrangements, the boys can be placed in $3!$ ways, and for each of these placements, the girls can be arranged in $3!$ ways, for a total of $(4)(3!)(3!)$.

That way of thinking about things might be useful if instead we want, for example, the number of arrangements that have a girl at each end.

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Consider the girls as $g_1, g_2$ and $g_3$. They are being arranged in the $3!$ ways as mentioned in the solution.

These 3 girls are taken as a single set, $g$ (suppose). Then the arrangements are to be from the set of $b_1, b_2, b_3$ and $g$.

Total 4 elements arranged in $4!$ ways, and $3!$ ways of girls, giving you $24*6 = 144$ ways.

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