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monroe17

Professor Bartlett teaches a class of 11 students. She has a visually impaired student, Louise, who must sit in the front row next to her tutor, who is also a member of this class. Assume that the front row has eight chairs, and the tutor must be seated to the right of Louise. How many different ways can professor Bartlett assign students to sit in the first row?

  • one year ago
  • one year ago

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  1. monroe17
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    @jim_thompson5910 help? :)

    • one year ago
  2. jim_thompson5910
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    A big key phrase here is that "the tutor must be seated to the right of Louise", so if T is the tutor and L is Louise, then you can only have LT and NOT TL So basically LT is one person since you can't a) separate the two AND b) you can't reorder it

    • one year ago
  3. monroe17
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    so 6 spaces left right

    • one year ago
  4. jim_thompson5910
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    So instead of 11 students to order, you really have 11-2+1 = 10 students to order since you're combining two students to form one "student'

    • one year ago
  5. monroe17
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    10P6 maybe?

    • one year ago
  6. jim_thompson5910
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    There are 8 chairs in the front, but 2 are taken up and combined into one "chair" so to speak. So there are really 8-2+1 = 7 "chairs" in the front row.

    • one year ago
  7. jim_thompson5910
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    So it's really 10 P 7

    • one year ago
  8. monroe17
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    so 10P7?

    • one year ago
  9. jim_thompson5910
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    you got it

    • one year ago
  10. monroe17
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    5040 ways?

    • one year ago
  11. jim_thompson5910
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    5040 is 7! or 7 P 7, so no

    • one year ago
  12. monroe17
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    whats 10P7 then?

    • one year ago
  13. jim_thompson5910
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    10 P 7 = (10!)/((10-7)!)

    • one year ago
  14. monroe17
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    604800= (10!)/(3!)?

    • one year ago
  15. jim_thompson5910
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    you got it

    • one year ago
  16. monroe17
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    finally thank you!

    • one year ago
  17. jim_thompson5910
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    you're welcome

    • one year ago
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