Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 11"
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case 1 blue: Total number of octahedrons with the condition = 0 | case 1 blue: Total number of octahedrons with the condition = 0 | ||
| − | case 2 blues: Total number of octahedrons with the condition = <math> | + | case 2 blues: Total number of octahedrons with the condition = <math>\binom{8}{2}-2\binom{4}{2}-2=14</math> |
| − | case 3 blues: Total number of octahedrons with the condition = <math>\binom{8}{3}-binom{4}{3}= | + | case 3 blues: Total number of octahedrons with the condition = <math>\binom{8}{3}-2\binom{4}{3}=48</math> |
| − | case 4 blues: Total number of octahedrons with the condition = <math>\binom{8}{4}-2=68</math> | + | case 4 blues: Total number of octahedrons with the condition = <math>\binom{8}{4}-2\binom{8}{4}=68</math> |
| − | case 5 blues: Total number of octahedrons with the condition = Total from case 3 = | + | case 5 blues: Total number of octahedrons with the condition = Total from case 3 = 48 |
| − | case 6 blues: Total number of octahedrons with the condition = Total from case 2 = | + | case 6 blues: Total number of octahedrons with the condition = Total from case 2 = 14 |
case 7 blues: Total number of octahedrons with the condition = Total from case 1 = 0 | case 7 blues: Total number of octahedrons with the condition = Total from case 1 = 0 | ||
| Line 22: | Line 22: | ||
case 8 blues: Total number of octahedrons with the condition = Total from case 0 = 0 | case 8 blues: Total number of octahedrons with the condition = Total from case 0 = 0 | ||
| − | Gran Total <math>= | + | Gran Total <math>=14+48+68+48+14=192</math> |
| + | Probability = <math>\frac{192}{2^8}=\frac{192}{256}=\frac{3}{4}</math> | ||
| + | <math>m=3</math>, <math>n=4</math>, <math>m+n=3+4=\boxed{7}</math> | ||
~Tomas Diaz. orders@tomasdiaz.com | ~Tomas Diaz. orders@tomasdiaz.com | ||
{{alternate solutions}} | {{alternate solutions}} | ||
Latest revision as of 17:40, 26 November 2023
Problem
Each face of an octahedron is randomly colored blue or red. A caterpillar is on a vertex of the octahedron and wants to get to the opposite vertex by traversing the edges. The probability that it can do so without traveling along an edge that is shared by two faces of the same color is 
, where 
and 
are relatively prime positive integers. Find 
.
Solution
case 0 blues: Total number of octahedrons with the condition = 0
case 1 blue: Total number of octahedrons with the condition = 0
case 2 blues: Total number of octahedrons with the condition = 
case 3 blues: Total number of octahedrons with the condition = 
case 4 blues: Total number of octahedrons with the condition = 
case 5 blues: Total number of octahedrons with the condition = Total from case 3 = 48
case 6 blues: Total number of octahedrons with the condition = Total from case 2 = 14
case 7 blues: Total number of octahedrons with the condition = Total from case 1 = 0
case 8 blues: Total number of octahedrons with the condition = Total from case 0 = 0
Gran Total 
Probability = 
, 
, 
~Tomas Diaz. orders@tomasdiaz.com
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
