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Problem on probability

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Problem on probability

by Pinku » Thu Jul 22, 2010 9:25 pm
Given that there are three couples, who are to be arranged in 6 seats. Find how many ways they can be arranged, such that husband and wife sit together?

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by 4GMAT_Mumbai » Thu Jul 22, 2010 9:46 pm
Let AB, PQ and XY be the three couples.

These 3 couples can be arranged in 3! ways.

Each couple can seat themselves in 2! ways in the two chairs allotted to them.

Hence, 3! * 2! * 2! * 2! = 48 ways.

By the way, is this an 'arrangement' question or a 'probability' question ?

If it is a probability question,

P(the event above) = 48 / 6!.

Hope this helps. Thanks.
Naveenan Ramachandran

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by outreach » Thu Jul 22, 2010 9:48 pm
let us combine husband and wife together.
both of them together will choose 2 seats....

each pair can choose 3(6/2) seats in 3!
each pair can be arranged in 2!

so total= 3! * (2!)^3
= 6 * 2^3
= 6 * 8
= 48
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