anonymous
  • anonymous
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?
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
@jim_thompson5910 help? :)
jim_thompson5910
  • jim_thompson5910
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
anonymous
  • anonymous
so 6 spaces left right

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jim_thompson5910
  • jim_thompson5910
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'
anonymous
  • anonymous
10P6 maybe?
jim_thompson5910
  • jim_thompson5910
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.
jim_thompson5910
  • jim_thompson5910
So it's really 10 P 7
anonymous
  • anonymous
so 10P7?
jim_thompson5910
  • jim_thompson5910
you got it
anonymous
  • anonymous
5040 ways?
jim_thompson5910
  • jim_thompson5910
5040 is 7! or 7 P 7, so no
anonymous
  • anonymous
whats 10P7 then?
jim_thompson5910
  • jim_thompson5910
10 P 7 = (10!)/((10-7)!)
anonymous
  • anonymous
604800= (10!)/(3!)?
jim_thompson5910
  • jim_thompson5910
you got it
anonymous
  • anonymous
finally thank you!
jim_thompson5910
  • jim_thompson5910
you're welcome

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