# Math

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Four couples have purchased ticket to the school play. Their sear are next to one another in a single row. If each couple sits side by side, how many seating arrangements are possible?

(I came up with 24. correct??)

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I count 64 posibilitys: p=person s=seat
p ssssssss
p ssssssss
p ssssssss
p ssssssss
p ssssssss
p ssssssss
p ssssssss
p ssssssss

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is that with the requirement that each couple sit side by side?

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no

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if the couples have to sit side by side then i count 32

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In the first seat: There are 8 possibilites for who sits there.

In the second seat: There is only 1 possibility (the other member of the couple).

In the third seat: There are 6 possibilites for who sits there.

In the fourth seat: There is only 1 possibility (the other member of the couple).

In the fifth seat: There are 4 possibilities for who sits there.

In the sixth seat: There is only 1 possibility (the other member of the couple).

In the seventh seat: There are 2 possibilities for who sits there.

In the eighth seat: There is only 1 possibility (the last member of the last couple).

Answer: 8*1*6*1*4*1*2*1 = 384 different possibilities!

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There are 384 possibilities.

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uhhhhhhhh are you sure cause that doesn't sound posible. especialy if the couples have to sit side by side

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i know u included the couples sitting side by side but still

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Yes, if the couples were not required to sit next to each other then the number of possibilities would be (8*7*6*5*4*3*2*1) = 40,320. I've taken Statistics at Georgia Tech and this was the fun part!!

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got it. Thanks so much.

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there are six ways each of them can sit next to each of them sit. five to sit next to them. six times six = thirtysix
five times five = 30 add them together you get 66

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