Difference between revisions of "2014 AMC 12A Problems/Problem 13"
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We can discern three cases. | We can discern three cases. | ||
| − | Case 1: Each room houses one guest. In this case, we have <math>5</math> guests to choose for the first room, <math>4</math> for the second, ..., for a total of <math>5!=120</math> assignments. | + | '''Case 1:''' Each room houses one guest. In this case, we have <math>5</math> guests to choose for the first room, <math>4</math> for the second, ..., for a total of <math>5!=120</math> assignments. |
| − | 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 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. | + | 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}=900</math> ways. | ||
In total, there are <math>120+1200+900=\boxed{2220}</math> assignments, or <math>\textbf{(B)}</math>. | In total, there are <math>120+1200+900=\boxed{2220}</math> assignments, or <math>\textbf{(B)}</math>. | ||
Revision as of 19:28, 7 February 2014
Problem
Solution
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 
.
