Here you choose 3 people to go into one of the 4 possible rooms. Then there are only 3 possibilities of where the remaining 3 people can go. Here you do the same thing except choosing 1 of the 2 remaining people to go into the other 3 rooms.Ĥ,1,1,0 C(6,4)*(4 rooms)*C(2,1)*(3 rooms)*(2 rooms) = 720Ĭhoose 3 of the 6 people to go into one of the 4 rooms (4 possibilities).
The remaining 2 people have to go into one of the other 3 rooms (3 possibilities). Possible ways include (4,2,0,0), (4,1,1,0), (3,3,0,0) and (3,1,1,1).įirst you choose 4 from the 6 people and multiply by the 4 possible rooms they can go into. There are only so many ways to arrange the 6 people since only 4 can fit in each room. So, subtract (4+72 = 76 ) from 4096 to get 4020.Ĭombination of n elements where you are choosing k elements from it: How many are there? “How many times may 2 people (that is the fifth and sixth) be assigned to 4 rooms?”Ĥ of those arrangements have all 6 people in 1 room (since it is all 6 people, we count this once).įor 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. To account for the restriction that 5 and 6 assignments of a room is not acceptable, we must subtract those combinations. So far, this assumes that rooms may be assigned any number of times (that is, unlimited replacement is allowed).
Now, if each of 6 persons may be assigned to one of the 4 rooms, there are 4*4*4*4*4*4 = 46 = 4096 possible assignments. 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 is counted only once. We assume that rooms are indistinguishable. We have already assumed that assigning 0 people to a room is o.k. You have a graph $G$ on ten vertices (the ten people in the room $ = 5$.(2) assign 6 people to 4 rooms but do not allow more than 4 people to a room. You can think of this as a problem in graph theory.
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