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 one year ago
At the head table at a banquet are seated two senators, two governors, and three mayors. Find the number of ways in which these seven people can be seated under the conditions described: A mayor is at each end and the senators are in consecutive seats
 one year ago
At the head table at a banquet are seated two senators, two governors, and three mayors. Find the number of ways in which these seven people can be seated under the conditions described: A mayor is at each end and the senators are in consecutive seats

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satellite73
 one year ago
Best ResponseYou've already chosen the best response.0dw:1361848754746:dw

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0lots of applications of the counting principle two choices for mayor

electronicz
 one year ago
Best ResponseYou've already chosen the best response.0ok so how will u do the senators?

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0oh i read it wrong, sorry

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0lets back up a bit

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0label the seats 1 to 7 then senators can be in seats 2, 3 3, 4 4, 5 5, 6 and they can switch chairs between them, so there are 8 possibilities for the senators

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0seat number 1 and 7 must contain mayors, so there are \(3\times 2=6\) choices for seat 1 and 7

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0then you have 3 chairs left to fill , and 3 people left to sit, so they are \(3\times 2=6\) choices for the remaining 3 chairs

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0multiply all these choices together, and unless i screwed up you get the right answer

electronicz
 one year ago
Best ResponseYou've already chosen the best response.0... the answer is 12

Mertsj
 one year ago
Best ResponseYou've already chosen the best response.1I do not see how the answer can only be 12.

electronicz
 one year ago
Best ResponseYou've already chosen the best response.0well i guess order doesn't matter i nthnis case

electronicz
 one year ago
Best ResponseYou've already chosen the best response.0yea its 12... hey can u just help me on this question? how many committees of four can be chosen from twelve students? how many of these will include a given student? how many will exclude a given student?

Mertsj
 one year ago
Best ResponseYou've already chosen the best response.1It's a comination problem so 12 choose 4

electronicz
 one year ago
Best ResponseYou've already chosen the best response.0yes what about the last two questions?

Mertsj
 one year ago
Best ResponseYou've already chosen the best response.1If we want to include a given student, we have already made one choice so we now must choose3 students from the remaining 11

Mertsj
 one year ago
Best ResponseYou've already chosen the best response.1So I think it would be 11 choose 3

Mertsj
 one year ago
Best ResponseYou've already chosen the best response.1I'm not really very good with probability.

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0you have to multiply all the choices together, as i wrote above

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0unless i screwed up, it should be \(8\times 6\times 6\)
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