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Word Problem Involving Probability

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Word Problem Involving Probability

by EbrahimHashem » Tue Mar 23, 2010 11:30 am
A rectangular table seats 4 people on each of two sides, with every person directly facing another person across the table. If eight people choose their seats at random, what is the probability that any two of them directly face each other?

A) 1/56

B) 1/8

C) 1/7

D) 15/56

E) 4/7

Enjoy it :D
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by newton9 » Tue Mar 23, 2010 12:14 pm
Total number of possible combinations is 8!.

And I select person-1 and person-2 who will sit opposite to each other. I can arrange them in 2 ways. And all the remaining 6 in 6! ways.

Person 1 and 2 can sit opposite in any of the 4 tables.

So total combinations are : 6! x 4 x 2.

That is how I arrived at C - 1/7.

Please give the OA.
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by thephoenix » Wed Mar 24, 2010 12:03 am
# OF COMBINATION for placing 8 people =8!

now a person can ca be seated in 8 ways
second person can be seated in only one way as he has only one seat available in front of the Ist person

rest 6 can be seated in 6 available seats in 6!

prob=8*6!/8!=1/7
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