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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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
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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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
so 6 spaces left right

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Other answers:

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

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