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by Rich@VeritasPrep » Tue Jun 29, 2010 7:42 pm
The arrangement has to be B-G-B-G-B-G-B

There are 4! = 24 ways to arrange the boys, and for each of these, there are 3! = 6 ways to arrange the girls. That's a total of 24*6 = 144 possible arrangements in which the boys and girls alternate.

With 7 total students, there are 7! = 5040 possible arrangements.

We're interested in 144 out of 5040, so the probability is:

[spoiler]144/5040 = 1/35[/spoiler]

Hope that helps!
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by Rahul@gurome » Tue Jun 29, 2010 7:47 pm
Probability is (No. of desired outcomes)/ (Total no. of outcomes).

First calculate the number of desired outcomes.
If 4 boys are seated in a row there are 3 alternate places between them.
The 4 boys can be arranged among themselves in 4! which is 24 ways and the 3 girls can be arranged in the 3 places in 3! or 6 ways.
So No. of desired outcomes is 24*6 = 144.
Also total 4+3 = 7 boys and girls can be arranged in 7! ways.
So required probability is 144/7! = 1/35.
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