anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
hi
anonymous
  • anonymous
|dw:1361848754746:dw|
anonymous
  • anonymous
lots of applications of the counting principle two choices for mayor

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More answers

anonymous
  • anonymous
ok so how will u do the senators?
anonymous
  • anonymous
oh i read it wrong, sorry
anonymous
  • anonymous
lets back up a bit
anonymous
  • anonymous
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
  • anonymous
seat number 1 and 7 must contain mayors, so there are \(3\times 2=6\) choices for seat 1 and 7
anonymous
  • anonymous
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
  • anonymous
multiply all these choices together, and unless i screwed up you get the right answer
anonymous
  • anonymous
... the answer is 12
anonymous
  • anonymous
@Mertsj ?
Mertsj
  • Mertsj
I do not see how the answer can only be 12.
Mertsj
  • Mertsj
Are you sure it is 12?
anonymous
  • anonymous
well i guess order doesn't matter i nthnis case
anonymous
  • anonymous
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
  • Mertsj
It's a comination problem so 12 choose 4
Mertsj
  • Mertsj
|dw:1361850098787:dw|
anonymous
  • anonymous
yes what about the last two questions?
Mertsj
  • Mertsj
If we want to include a given student, we have already made one choice so we now must choose3 students from the remaining 11
anonymous
  • anonymous
ok
Mertsj
  • Mertsj
So I think it would be 11 choose 3
Mertsj
  • Mertsj
I'm not really very good with probability.
anonymous
  • anonymous
you have to multiply all the choices together, as i wrote above
anonymous
  • anonymous
unless i screwed up, it should be \(8\times 6\times 6\)

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