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There are 5 boys and 4 girls in my class. All of them are distinguishable.

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There are 5 boys and 4 girls in my class. All of them are distinguishable.

In how many ways can they be seated in a row of 9 chairs such that at least 3 girls are all next to each other?

Mar 5, 2015

#1
+5

ok

how many ways can the 3 girls sit next to each other ?   4P3 = 24

Now lets take just one of these 24 permutations and put them together with the others.

I will take the 3 girls as one (siamese triplets) + one more girl +5boys = 7  so 7! permutations.

So altogether I think that there will be   24*7! permutations

$${\mathtt{24}}{\mathtt{\,\times\,}}{\mathtt{7}}{!} = {\mathtt{120\,960}}$$ I think that is correct.

Having that 4th girl is throwing me a bit, but I think it is ok.

Mar 6, 2015

#1
+5

ok

how many ways can the 3 girls sit next to each other ?   4P3 = 24

Now lets take just one of these 24 permutations and put them together with the others.

I will take the 3 girls as one (siamese triplets) + one more girl +5boys = 7  so 7! permutations.

So altogether I think that there will be   24*7! permutations

$${\mathtt{24}}{\mathtt{\,\times\,}}{\mathtt{7}}{!} = {\mathtt{120\,960}}$$ I think that is correct.

Having that 4th girl is throwing me a bit, but I think it is ok.

Melody Mar 6, 2015