Problem on probability

[Pinku]

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?

[4GMAT_Mumbai]

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.

[outreach]

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