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monroe17

  • 2 years ago

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?

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  1. monroe17
    • 2 years ago
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    @jim_thompson5910 help? :)

  2. jim_thompson5910
    • 2 years ago
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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

  3. monroe17
    • 2 years ago
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    so 6 spaces left right

  4. jim_thompson5910
    • 2 years ago
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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'

  5. monroe17
    • 2 years ago
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    10P6 maybe?

  6. jim_thompson5910
    • 2 years ago
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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.

  7. jim_thompson5910
    • 2 years ago
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    So it's really 10 P 7

  8. monroe17
    • 2 years ago
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    so 10P7?

  9. jim_thompson5910
    • 2 years ago
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    you got it

  10. monroe17
    • 2 years ago
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    5040 ways?

  11. jim_thompson5910
    • 2 years ago
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    5040 is 7! or 7 P 7, so no

  12. monroe17
    • 2 years ago
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    whats 10P7 then?

  13. jim_thompson5910
    • 2 years ago
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    10 P 7 = (10!)/((10-7)!)

  14. monroe17
    • 2 years ago
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    604800= (10!)/(3!)?

  15. jim_thompson5910
    • 2 years ago
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    you got it

  16. monroe17
    • 2 years ago
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    finally thank you!

  17. jim_thompson5910
    • 2 years ago
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    you're welcome

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