2014 AMC 12A Problem 13

Question

A fancy bed and breakfast inn has

5

rooms, each with a distinctive color-coded decor. One day

5

friends arrive to spend the night. There are no other guests that night. The friends can room in any combination they wish, but with no more than

2

friends per room. In how many ways can the innkeeper assign the guests to the rooms?

2100

2220

3000

3120

3125

Split into cases by the room occupancies: all singles, one pair, or two pairs

Count each case as (ways to form the groups) times (ways to place the groups into distinct rooms)

All singles: assign

5

friends to

5

rooms in

5!=120

ways.

One pair: choose the pair in

\binom52=10

ways, then place the

4

groups into rooms in

5\cdot4\cdot3\cdot2=120

ways, giving

10\cdot120=1200.

Two pairs: choose the solo friend in

5

ways and split the rest into two pairs in

3

ways (

15

groupings), then place the

3

groups into rooms in

5\cdot4\cdot3=60

ways, giving

15\cdot60=900.

The total is

120+1200+900=2220.

Thus, the correct answer is B.

Problem 13 in other years