Difference between revisions of "2012 AMC 10B Problems/Problem 25"
Sorcerer76 (talk | contribs) (→Solution) |
Username222 (talk | contribs) |
||
Line 95: | Line 95: | ||
Finally, from each of the first and second green arrows, there is respectively <math>2</math> way to get to the first orange arrow; from each of the third and the fourth green arrows, there are <math>3</math> ways to get to the first orange arrow. Therefore there are <math>120 \cdot (2+2+3+3) = 1200</math> ways to get to each of the orange arrows, hence <math>2400</math> ways to get to the point <math>B</math>. <math>\framebox{E}</math> | Finally, from each of the first and second green arrows, there is respectively <math>2</math> way to get to the first orange arrow; from each of the third and the fourth green arrows, there are <math>3</math> ways to get to the first orange arrow. Therefore there are <math>120 \cdot (2+2+3+3) = 1200</math> ways to get to each of the orange arrows, hence <math>2400</math> ways to get to the point <math>B</math>. <math>\framebox{E}</math> | ||
+ | |||
+ | |||
+ | == See Also == | ||
+ | |||
+ | {{AMC12 box|year=2012|ab=B|num-b=24|num-a=26}} |
Revision as of 23:23, 12 January 2013
- The following problem is from both the 2012 AMC 12B #22 and 2012 AMC 10B #25, so both problems redirect to this page.
A bug travels from A to B along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there?
Solution
There is way to get to any of the red arrows. From the first red arrow, there are ways to get to each of the first and the second blue arrows; from the second red arrow, there are ways to get to each of the first and the second blue arrows. So there are in total ways to get to each of the blue arrows.
From each of the first and second blue arrows, there are respectively ways to get to each of the first and the second green arrows; from each of the third and the fourth blue arrows, there are respectively ways to get to each of the first and the second green arrows. Therefore there are in total ways to get to each of the green arrows.
Finally, from each of the first and second green arrows, there is respectively way to get to the first orange arrow; from each of the third and the fourth green arrows, there are ways to get to the first orange arrow. Therefore there are ways to get to each of the orange arrows, hence ways to get to the point .
See Also
2012 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Problem 26 |
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 |