Perms and Combs Qs (1 Viewer)

[mikey1312]

Four girls and five boys are to sit in a line. Find the number of permutations if:
a) 2 girls (only 2, not 3 nor 2 pairs) sit together.
b) 3 girls (only 3, not 4) sit together.

 

[Carrotsticks]

A little clarification needed. When you say 2 girls sit together, do you mean ANY two girls, or two PARTICULAR girls?

Because although subtle, it yields different results.

 

[mikey1312]

Any 2 girls.

 

[bleakarcher]

Four girls and five boys are to sit in a line. Find the number of permutations if:
a) 2 girls (only 2, not 3 nor 2 pairs) sit together.
b) 3 girls (only 3, not 4) sit together.

Is the answer 63360 permutations for the first part?

 

[RealiseNothing]

Is the answer 63360 permutations for the first part?

I got 28800, but the wording is really ambiguous so not sure.

 

[bleakarcher]

I got 28800, but the wording is really ambiguous so not sure.

Just revised my logic and I'm pretty sure it's absolutely fucked.

 

[Sy123]

Hmm, how did you do it?
I got 14400 by doing

2!2!7!-2!2!2!6!

The logic is that, the first part of my answer (the one with 2 of the 2!), is the number of arrangements possible, that group girls together in pairs, disregarding the possibility that the 2 pairs of girls could be placed next to each other. Then I take away the number of arrangments possible that would group together these girls together. The 6 factorial is bundling together 4 girls into one bundle, and 5 other boys, however in this 4 girl bundle, it can only be arranged in 2! ways, because theoretically, these pairs of girls cannot be split up, however their position of the pairs can move, however the position of girls within their own pairs can change, so that is where the other 2! comes from.

 

[RealiseNothing]

I just did it again and got this:

The brackets "(only 2, not 3 nor 2 pairs)" makes me think we have to put 2 girls together, and that no other girls can sit together (they must be next to boys).

Let two girls be one element.

Now we have essentially 3 girls and 5 boys, so let's arrange them so that none of these 3 sit together. This is actually sum of the first 4 triangular numbers (1+3+6+10 = 20). They can then be arranged in 4! different ways, so 4! times 20 which is 480. This is the arrangements of all girls. Since boys don't have a restriction, we can seat them in the remaining 5 spots in 5! ways, which is 120 . Multiply the arrangements together and you get 480x120.

So the answer should be 57600. But honestly this question has the worst wording and I doubt my answer is right.