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How many ways can 7 people be placed into 10 rooms, if (only) 2 of them can’t share a room with anyone?

I'm not sure how to go about this, mostly because of the "share a room" bit. I'm thinking I should calculate the number of ways 2 people can be placed into 10 rooms, then the number of ways the remaining 5 people can be placed into the 8 rooms left, and multiply the two figures together. Am I on the right track?

Also, does sharing rooms mean "with replacement"?

L.Price
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  • Are there two particular people that won't share? If so, then your approach is right. Figure out how many ways you can place those two in $10$ rooms. Then figure out how many ways you can place the other five in $8$ rooms, and then multiply. – mjqxxxx Oct 23 '13 at 23:40
  • Yes, you are on the right track. – Newb Oct 23 '13 at 23:44

1 Answers1

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Community wiki answer so the question can be marked as answered:

As noted in the comments, you were on the right track. There are $\binom{10}2$ choices for the two rooms for the hermites, and then $8^5$ ways to select one of the remaining $8$ rooms for each of the remaining $5$ people, for a total of $\binom{10}2\cdot8^5=1474560$.

joriki
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  • Shouldn't it be $10\cdot9$ instead of $10\choose2$? – Barry Cipra Dec 22 '17 at 03:20
  • I'm a newb so I could be totally off but I also think it should be $10*9$. Because by doing (10 choose 2) you are not distinguishing between the selections of the two loners. For example the first loner choosing room 1 and second loner choosing room 2 is the same as the first loner choosing room 2 and the second loner choosing room 1. This should count as 2 different configurations, however by doing (10 choose 2) it would count as 1... please correct me if I'm wrong. – HJ_beginner Dec 22 '17 at 09:09