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hi

|dw:1361848754746:dw|

lots of applications of the counting principle
two choices for mayor

ok so how will u do the senators?

oh i read it wrong, sorry

lets back up a bit

seat number 1 and 7 must contain mayors, so there are \(3\times 2=6\) choices for seat 1 and 7

multiply all these choices together, and unless i screwed up you get the right answer

... the answer is 12

I do not see how the answer can only be 12.

Are you sure it is 12?

well i guess order doesn't matter i nthnis case

It's a comination problem so 12 choose 4

|dw:1361850098787:dw|

yes what about the last two questions?

ok

So I think it would be 11 choose 3

I'm not really very good with probability.

you have to multiply all the choices together, as i wrote above

unless i screwed up, it should be \(8\times 6\times 6\)