Difference between revisions of "2005 AIME II Problems/Problem 2"
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== Solution == | == Solution == | ||
=== Solution 1 === | === Solution 1 === | ||
| − | Use [[ | + | Use [[construction]]. 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 1: <math>\frac{9 \cdot 6 \cdot 3}{9 \cdot 8 \cdot 7} = \frac{9}{28}</math> | ||
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*Person 3: One roll of each type is left, so the probability here is <math>1</math>. | *Person 3: One roll of each type is left, so the probability here is <math>1</math>. | ||
| − | Our answer is thus <math>\frac{9}{28} \cdot \frac{2}{5} = \frac{9}{70}</math>, and <math>m + n = 79</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 === | === Solution 2 === | ||
Revision as of 17:13, 22 July 2017
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 construction. 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
