Difference between revisions of "2014 AMC 12A Problems/Problem 13"
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\textbf{(E) }3125\qquad</math> | \textbf{(E) }3125\qquad</math> | ||
| − | ==Solution== | + | ==Solution 1== |
We can discern three cases. | We can discern three cases. | ||
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In total, there are <math>120+1200+900=2220</math> assignments, or <math>\boxed{\textbf{(B)}}</math>. | In total, there are <math>120+1200+900=2220</math> assignments, or <math>\boxed{\textbf{(B)}}</math>. | ||
| − | (Solution by | + | (Solution by AwesomeToad16) |
| + | |||
| + | ==Solution 2== | ||
| + | |||
| + | We can work in reverse by first determining the number of combinations in which there are more than <math>2</math> friends in at least one room. There are three cases: | ||
| + | |||
| + | '''Case 1:''' Three friends are in one room. Since there are <math>5</math> possible rooms in which this can occur, we are choosing three friends from the five, and the other two friends can each be in any of the four remaining rooms, | ||
| + | there are <math>5\cdot\binom{5}{3}\cdot4\cdot4 = 800</math> possibilities. | ||
| + | |||
| + | '''Case 2:''' Four friends are in one room. Again, there are <math>5</math> possible rooms, we are choosing four of the five friends, and the other one can be in any of the other four rooms, so there are <math>5\cdot\binom{5}{4}\cdot4= 100</math> possibilities. | ||
| + | |||
| + | '''Case 3:''' Five friends are in one room. There are <math>5</math> possible rooms in which this can occur, so there are <math>5</math> possibilities. | ||
| + | |||
| + | Since there are <math>5^5 = 3125</math> possible combinations of the friends, the number fitting the given criteria is <math>3125 - (800+100+5) = \boxed{2220}</math>. | ||
| + | |||
| + | == Video Solution by OmegaLearn == | ||
| + | https://youtu.be/kxgUdv_L-ys?t=13 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC12 box|year=2014|ab=A|num-b=12|num-a=14}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 03:49, 4 November 2022
Problem
A fancy bed and breakfast inn has 
rooms, each with a distinctive color-coded decor. One day 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 
friends per room. In how many ways can the innkeeper assign the guests to the rooms?
Solution 1
We can discern three cases.
Case 1: Each room houses one guest. In this case, we have guests to choose for the first room, 
for the second, ..., for a total of 
assignments.
Case 2: Three rooms house one guest; one houses two. We have 
ways to choose the three rooms with 
guest, and 
to choose the remaining one with . There are 
ways to place guests in the first three rooms, with the last two residing in the two-person room, for a total of 
ways.
Case 3: Two rooms house two guests; one houses one. We have 
to choose the two rooms with two people, and 
to choose one remaining room for one person. Then there are choices for the lonely person, and 
for the two in the first two-person room. The last two will stay in the other two-room, so there are 
ways.
In total, there are 
assignments, or 
.
(Solution by AwesomeToad16)
Solution 2
We can work in reverse by first determining the number of combinations in which there are more than friends in at least one room. There are three cases:
Case 1: Three friends are in one room. Since there are possible rooms in which this can occur, we are choosing three friends from the five, and the other two friends can each be in any of the four remaining rooms,
there are 
possibilities.
Case 2: Four friends are in one room. Again, there are possible rooms, we are choosing four of the five friends, and the other one can be in any of the other four rooms, so there are 
possibilities.
Case 3: Five friends are in one room. There are possible rooms in which this can occur, so there are possibilities.
Since there are 
possible combinations of the friends, the number fitting the given criteria is 
.
Video Solution by OmegaLearn
https://youtu.be/kxgUdv_L-ys?t=13
~ pi_is_3.14
See Also
| 2014 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 12 |
Followed by Problem 14 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
