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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
 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
 one year ago

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satellite73Best ResponseYou've already chosen the best response.0
dw:1361848754746:dw
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
lots of applications of the counting principle two choices for mayor
 one year ago

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

satellite73Best ResponseYou've already chosen the best response.0
oh i read it wrong, sorry
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
lets back up a bit
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
label 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
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
seat number 1 and 7 must contain mayors, so there are \(3\times 2=6\) choices for seat 1 and 7
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
then 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
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
multiply all these choices together, and unless i screwed up you get the right answer
 one year ago

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

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

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

electroniczBest ResponseYou've already chosen the best response.0
yea 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?
 one year ago

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

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

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

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

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

satellite73Best ResponseYou've already chosen the best response.0
you have to multiply all the choices together, as i wrote above
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
unless i screwed up, it should be \(8\times 6\times 6\)
 one year ago
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