electronicz
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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electronicz
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hi
anonymous
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|dw:1361848754746:dw|
anonymous
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lots of applications of the counting principle
two choices for mayor
electronicz
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ok so how will u do the senators?
anonymous
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oh i read it wrong, sorry
anonymous
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lets back up a bit
anonymous
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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
anonymous
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seat number 1 and 7 must contain mayors, so there are \(3\times 2=6\) choices for seat 1 and 7
anonymous
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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
anonymous
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multiply all these choices together, and unless i screwed up you get the right answer
electronicz
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... the answer is 12
electronicz
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@Mertsj
?
Mertsj
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I do not see how the answer can only be 12.
Mertsj
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Are you sure it is 12?
electronicz
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well i guess order doesn't matter i nthnis case
electronicz
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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?
Mertsj
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It's a comination problem so 12 choose 4
Mertsj
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electronicz
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yes what about the last two questions?
Mertsj
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If we want to include a given student, we have already made one choice so we now must choose3 students from the remaining 11
electronicz
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ok
Mertsj
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So I think it would be 11 choose 3
Mertsj
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I'm not really very good with probability.
anonymous
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you have to multiply all the choices together, as i wrote above
anonymous
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unless i screwed up, it should be \(8\times 6\times 6\)