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"?
Best Answer
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 $\binom{10}2\cdot8^5=1474560$.