SOLUTION: Nine guests are invited for dinner. If there are 2 "ends" to the table, how many different pairs can be seated on the ends? Assume the seating is random (a "couple" can be split u

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Question 460931: Nine guests are invited for dinner. If there are 2 "ends" to the table, how many different pairs can be seated on the ends?
Assume the seating is random (a "couple" can be split up) and the same pair at either end is still only one pair.
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Nine guests are invited for dinner. If there are 2 "ends" to the table, how many different pairs can be seated on the ends?
Assume the seating is random (a "couple" can be split up) and the same pair at either end is still only one pair.
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# of ways to pick a "pair": 9C2 = 36
# of ways to pick an "end": 2
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# of different "pairs" on the "ends" = 2*36 = 72
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Cheers,
Stan H.