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anonymous
 one year ago
how many ways can a music teacher arrange 24 students in 3 rows?
anonymous
 one year ago
how many ways can a music teacher arrange 24 students in 3 rows?

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welshfella
 one year ago
Best ResponseYou've already chosen the best response.1First I guess you must have 8 students in each row right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes that would be correct..

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1And do you know how to calculate the number of ways 8 people can be arranged in a row?

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1Well you will need a calculator to do it Have you heard of factorial 8. : 8!

welshfella
 one year ago
Best ResponseYou've already chosen the best response.18! = 8*7*6*5*4*3*2*1

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1And as there is also 3 rows the calculations is 3! * 8!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0In how many ways can a teacher arrange 5 students in the front row of a classroom with a total of 20 students

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@welshfella I think there's more to the first problem, but I could be wrong. We're not explicitly told whether the same 8 people are seated in any given row. I think this means we have \(8!\dbinom{24}8\) (or \({}_{24}P_8\)) ways of seating the first 8 students in one row, then \(8!\dbinom{16}8\) in the next, and finally \(8!\dbinom88\) in the third. So in my mind there may be as many as \((8!)^3\dbinom{24}8\dbinom{16}8\) ways of seating 24 students in 3 rows.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Or, alternatively, you can write this as \((3!)^8\dbinom{24}3\dbinom{21}3\cdots\dbinom63\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This assuming order matters, which by the context I think we can assume it does.
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