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electronicz

  • 3 years 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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  1. electronicz
    • 3 years ago
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    hi

  2. anonymous
    • 3 years ago
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    |dw:1361848754746:dw|

  3. anonymous
    • 3 years ago
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    lots of applications of the counting principle two choices for mayor

  4. electronicz
    • 3 years ago
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    ok so how will u do the senators?

  5. anonymous
    • 3 years ago
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    oh i read it wrong, sorry

  6. anonymous
    • 3 years ago
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    lets back up a bit

  7. anonymous
    • 3 years ago
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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

  8. anonymous
    • 3 years ago
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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

  9. anonymous
    • 3 years ago
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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

  10. anonymous
    • 3 years ago
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    multiply all these choices together, and unless i screwed up you get the right answer

  11. electronicz
    • 3 years ago
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    ... the answer is 12

  12. electronicz
    • 3 years ago
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    @Mertsj ?

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

  14. Mertsj
    • 3 years ago
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    Are you sure it is 12?

  15. electronicz
    • 3 years ago
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    well i guess order doesn't matter i nthnis case

  16. electronicz
    • 3 years ago
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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?

  17. Mertsj
    • 3 years ago
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    It's a comination problem so 12 choose 4

  18. Mertsj
    • 3 years ago
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    |dw:1361850098787:dw|

  19. electronicz
    • 3 years ago
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    yes what about the last two questions?

  20. Mertsj
    • 3 years ago
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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

  21. electronicz
    • 3 years ago
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    ok

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

  23. Mertsj
    • 3 years ago
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    I'm not really very good with probability.

  24. anonymous
    • 3 years ago
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    you have to multiply all the choices together, as i wrote above

  25. anonymous
    • 3 years ago
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    unless i screwed up, it should be \(8\times 6\times 6\)

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