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You are sorting assigning 6 people, A, B, C, D, E and F, into 3 different hotel rooms. How many ways can they be sorted such that A is in the same room with C, and B is not in the same room with D? (Some hotel rooms may be empty.)

7C5 * 2! * 5! - 6C4 * 2! * 2! *4!

=5040-1440

=3600

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    I’ve not been able to figure out what you were thinking, so I don’t know just where you went astray. What are the $7$ things from which you’re choosing $5$?2012-12-09
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    I am confused with the placing distinct object into identical container(hotels) and the container maybe empty.2012-12-09
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    First of all, A and C placed in the same room Then, we count A and C as one object AC , B , D , E ,F to place them in 3 hotels it should be 7C5 right?2012-12-09
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    But the rooms aren’t identical. Were you perhaps thinking of [this method](http://en.wikipedia.org/wiki/Stars_and_bars_%28probability%29) that is used when the **objects** are identical and the containers are not? It doesn’t apply here, because the objects (people) are not identical.2012-12-09

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