Problem on probability
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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.
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
4GMAT, Dadar(W) & Ghatkopar(W), Mumbai
4GMAT, Dadar(W) & Ghatkopar(W), Mumbai
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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
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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