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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

  • one year ago
  • one year ago

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  1. electronicz
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    hi

    • one year ago
  2. satellite73
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    |dw:1361848754746:dw|

    • one year ago
  3. satellite73
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    lots of applications of the counting principle two choices for mayor

    • one year ago
  4. electronicz
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    ok so how will u do the senators?

    • one year ago
  5. satellite73
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    oh i read it wrong, sorry

    • one year ago
  6. satellite73
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    lets back up a bit

    • one year ago
  7. satellite73
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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

    • one year ago
  8. satellite73
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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

    • one year ago
  9. satellite73
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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

    • one year ago
  10. satellite73
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    multiply all these choices together, and unless i screwed up you get the right answer

    • one year ago
  11. electronicz
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    ... the answer is 12

    • one year ago
  12. electronicz
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    @Mertsj ?

    • one year ago
  13. Mertsj
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    I do not see how the answer can only be 12.

    • one year ago
  14. Mertsj
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    Are you sure it is 12?

    • one year ago
  15. electronicz
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    well i guess order doesn't matter i nthnis case

    • one year ago
  16. 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?

    • one year ago
  17. Mertsj
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    It's a comination problem so 12 choose 4

    • one year ago
  18. Mertsj
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    |dw:1361850098787:dw|

    • one year ago
  19. electronicz
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    yes what about the last two questions?

    • one year ago
  20. 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

    • one year ago
  21. electronicz
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    ok

    • one year ago
  22. Mertsj
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    So I think it would be 11 choose 3

    • one year ago
  23. Mertsj
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    I'm not really very good with probability.

    • one year ago
  24. satellite73
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    you have to multiply all the choices together, as i wrote above

    • one year ago
  25. satellite73
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    unless i screwed up, it should be \(8\times 6\times 6\)

    • one year ago
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