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Difference between revisions of "2014 AMC 12A Problems/Problem 13"

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'''Case 2:''' Three rooms house one guest; one houses two. We have <math>\binom{5}{3}</math> ways to choose the three rooms with <math>1</math> guest, and <math>\binom{2}{1}</math> to choose the remaining one with <math>2</math>. There are <math>5\cdot4\cdot3</math> ways to place guests in the first three rooms, with the last two residing in the two-person room, for a total of <math>\binom{5}{3}\binom{2}{1}\cdot5\cdot4\cdot3=1200</math> ways.
 
'''Case 2:''' Three rooms house one guest; one houses two. We have <math>\binom{5}{3}</math> ways to choose the three rooms with <math>1</math> guest, and <math>\binom{2}{1}</math> to choose the remaining one with <math>2</math>. There are <math>5\cdot4\cdot3</math> ways to place guests in the first three rooms, with the last two residing in the two-person room, for a total of <math>\binom{5}{3}\binom{2}{1}\cdot5\cdot4\cdot3=1200</math> ways.
  
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'''Case 3:''' Two rooms house two guests; one houses one. We have <math>\binom{5}{2}</math> to choose the two rooms with two people, and <math>\binom{3}{1}</math> to choose one remaining room for one person. Then there are <math>5</math> choices for the lonely person, and <math>\binom{4}{2}</math> for the two in the first two-person room. The last two will stay in the other two-room, so there are <math>\binom{5}{2}\binom{3}{1}\cdot5\cdot\binom{4}{2}=1800</math> ways.
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'''Case 3:''' Two rooms house two guests; one houses one. We have <math>\binom{5}{2}</math> to choose the two rooms with two people, and <math>\binom{3}{1}</math> to choose one remaining room for one person. Then there are <math>5</math> choices for the lonely person, and <math>\binom{4}{2}</math> for the two in the first two-person room. The last two will stay in the other two-room, so there are <math>\binom{5}{2}\binom{3}{1}\cdot5\cdot\binom{4}{2}=900</math> ways.
  
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In total, there are <math>120+1200+1800=3120</math> assignments, or <math>\boxed{\textbf{(B)}}</math>.
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In total, there are <math>120+1200+900=2220</math> assignments, or <math>\boxed{\textbf{(B)}}</math>.
  
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(Solution by Allamsetty2)
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(Solution by AwesomeToad16)
  
 
==Solution 2==
 
==Solution 2==
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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>.
 
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>.
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== Video Solution by OmegaLearn ==
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https://youtu.be/kxgUdv_L-ys?t=13
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~ pi_is_3.14
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2014|ab=A|num-b=12|num-a=14}}
 
{{AMC12 box|year=2014|ab=A|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 03:49, 4 November 2022

Problem

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?

$\textbf{(A) }2100\qquad \textbf{(B) }2220\qquad \textbf{(C) }3000\qquad \textbf{(D) }3120\qquad \textbf{(E) }3125\qquad$

Solution 1

We can discern three cases.

Case 1: Each room houses one guest. In this case, we have $5$ guests to choose for the first room, $4$ for the second, ..., for a total of $5!=120$ assignments.

Case 2: Three rooms house one guest; one houses two. We have $\binom{5}{3}$ ways to choose the three rooms with $1$ guest, and $\binom{2}{1}$ to choose the remaining one with $2$. There are $5\cdot4\cdot3$ ways to place guests in the first three rooms, with the last two residing in the two-person room, for a total of $\binom{5}{3}\binom{2}{1}\cdot5\cdot4\cdot3=1200$ ways.

Case 3: Two rooms house two guests; one houses one. We have $\binom{5}{2}$ to choose the two rooms with two people, and $\binom{3}{1}$ to choose one remaining room for one person. Then there are $5$ choices for the lonely person, and $\binom{4}{2}$ for the two in the first two-person room. The last two will stay in the other two-room, so there are $\binom{5}{2}\binom{3}{1}\cdot5\cdot\binom{4}{2}=900$ ways.

In total, there are $120+1200+900=2220$ assignments, or $\boxed{\textbf{(B)}}$.

(Solution by AwesomeToad16)

Solution 2

We can work in reverse by first determining the number of combinations in which there are more than $2$ friends in at least one room. There are three cases:

Case 1: Three friends are in one room. Since there are $5$ 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 $5\cdot\binom{5}{3}\cdot4\cdot4 = 800$ possibilities.

Case 2: Four friends are in one room. Again, there are $5$ 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 $5\cdot\binom{5}{4}\cdot4= 100$ possibilities.

Case 3: Five friends are in one room. There are $5$ possible rooms in which this can occur, so there are $5$ possibilities.

Since there are $5^5 = 3125$ possible combinations of the friends, the number fitting the given criteria is $3125 - (800+100+5) = \boxed{2220}$.

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. AMC logo.png

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