Question

In a hotel four rooms are available. Six persons are to be accommodated in these four rooms in such a way that each of these rooms contains at least one person and at most two people. Then the no. of all possible ways in which this can be done is?

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COMBINATIONS

Kayla K. asked • 08/20/16

4 rooms can accommodate a maximum of 4 people. How many ways can 6 people be accommodated in these rooms?

Combinations question - answer is 4020 but I need to know how to get there

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David W. answered • 08/20/16

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There are two ways to view this problem:

(1) assign 4 rooms to 6 people, but do not allow a room to have more than 4 people (that is, each room may only be assigned 0 to 4 times); and

(2) assign 6 people to 4 rooms but do not allow more than 4 people to a room.

We have already assumed that assigning 0 people to a room is o.k. We assume that rooms are indistinguishable [a.k.a., "with replacement"]. We must also assume that it matters which people are assigned to a room (that is, A-B-C is different than B-C-D), but that a given set of people [say {A,B,C} in any order] is counted only once.

Now, if each of 6 persons may be assigned to one of the 4 rooms [the first part of view (2)], there are 4*4*4*4*4*4 = 46 = 4096 possible assignments. So far, this assumes that rooms may be assigned any number of times (that is, unlimited replacement is allowed).

To account for the restriction that 5 and 6 assignments of a room is not acceptable, we must subtract those combinations. How many are there? “How many times may 2 people (that is the fifth and sixth) be assigned to 4 rooms?”

4 of those arrangements have all 6 people in 1 room (since it is all 6 people, we count this once).

For a 5-1 arrangement in 2 of the 4 rooms, we select the 1 person (6 people choices), then assign that person to 1 of 4 rooms, then assign the other 5 people (as a group) to one of the remaining 3 rooms. That makes 6*4*3 = 72 combinations.

So, subtract (4+72 = 76 ) from 4096 to get 4020.

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