Difference between revisions of "2005 AIME II Problems/Problem 2"
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== Problem == | == Problem == | ||
| − | A hotel packed breakfast for each of three guests. Each breakfast should have consisted of three types of rolls, one each of nut, cheese, and fruit rolls. The preparer wrapped each of the nine rolls and once wrapped, the rolls were indistinguishable from one another. She then randomly put three rolls in a bag for each of the guests. Given that the probability each guest got one roll of each type is <math> \frac mn, </math> where <math> m </math> and <math> n </math> are relatively prime | + | A hotel packed breakfast for each of three guests. Each breakfast should have consisted of three types of rolls, one each of nut, cheese, and fruit rolls. The preparer wrapped each of the nine rolls and once wrapped, the rolls were indistinguishable from one another. She then randomly put three rolls in a bag for each of the guests. Given that the [[probability]] each guest got one roll of each type is <math> \frac mn, </math> where <math> m </math> and <math> n </math> are [[relatively prime]] [[integer]]s, find <math> m+n. </math> |
| + | __TOC__ | ||
== Solution == | == Solution == | ||
| − | + | === Solution 1 === | |
| − | == | + | Use [[casework]]. We need only calculate the probability the first and second person all get a roll of each type, since then the rolls for the third person are determined. |
| − | * | + | *Person 1: <math>\frac{9 \cdot 6 \cdot 3}{9 \cdot 8 \cdot 7} = \frac{9}{28}</math> |
| − | + | *Person 2: <math>\frac{6 \cdot 4 \cdot 2}{6 \cdot 5 \cdot 4} = \frac 25</math> | |
| − | + | *Person 3: One roll of each type is left, so the probability here is <math>\displaystyle 1</math>. | |
| + | |||
| + | Our answer is thus <math>\frac{9}{28} \cdot \frac{2}{5} = \frac{9}{70}</math>, and <math>m + n = 79</math>. | ||
| + | |||
| + | === Solution 2 === | ||
| + | Call the three different type of rolls as A, B, and C. We need to arrange 3As, 3Bs, and 3Cs in a string such that A, B, and C appear in the first three, second three, etc. This can occur in <math>\left(\frac{3!}{1!1!1!}\right)^3 = 6^3 = 216</math> different manners. The total number of possible strings is <math>\frac{9!}{3!3!3!} = 1680</math>. The solution is therefore <math>\frac{216}{1680} = \frac{9}{70}</math>, and <math>m + n = 79</math>. | ||
| + | |||
| + | == See also == | ||
| + | {{AIME box|year=2005|n=II|num-b=1|num-a=3}} | ||
| + | |||
| + | [[Category:Intermediate Combinatorics Problems]] | ||
Revision as of 19:41, 21 March 2007
Problem
A hotel packed breakfast for each of three guests. Each breakfast should have consisted of three types of rolls, one each of nut, cheese, and fruit rolls. The preparer wrapped each of the nine rolls and once wrapped, the rolls were indistinguishable from one another. She then randomly put three rolls in a bag for each of the guests. Given that the probability each guest got one roll of each type is 
where 
and 
are relatively prime integers, find 
Solution
Solution 1
Use casework. We need only calculate the probability the first and second person all get a roll of each type, since then the rolls for the third person are determined.
- Person 1:
- Person 2:
- Person 3: One roll of each type is left, so the probability here is
.
.Our answer is thus 
, and 
.
Solution 2
Call the three different type of rolls as A, B, and C. We need to arrange 3As, 3Bs, and 3Cs in a string such that A, B, and C appear in the first three, second three, etc. This can occur in 
different manners. The total number of possible strings is 
. The solution is therefore 
, and .
See also
| 2005 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
