Difference between revisions of "2014 AIME I Problems/Problem 11"
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<cmath>{6\choose 3} = 20 (*2)</cmath> | <cmath>{6\choose 3} = 20 (*2)</cmath> | ||
− | Thus, there are 800 + 840 + 480 + 40 = 2160 cases. Because there | + | Thus, there are 800 + 840 + 480 + 40 = 2160 cases. Because there are <math>4^6</math> total cases, the answer is <math>\frac{2160}{4^6} = \frac{135}{256}</math>, so the answer is <math>\boxed{391}.</math> |
== See also == | == See also == | ||
{{AIME box|year=2014|n=I|num-b=10|num-a=12}} | {{AIME box|year=2014|n=I|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 21:41, 22 March 2014
Problem 11
A token starts at the point of an
-coordinate grid and then makes a sequence of six moves. Each move is 1 unit in a direction parallel to one of the coordinate axes. Each move is selected randomly from the four possible directions and independently of the other moves. The probability the token ends at a point on the graph of
is
, where
and
are relatively prime positive integers. Find
.
Solution (casework)
We have 4 possible moves: U, D, R, and L. The total number of paths that could be taken is , or
. There are 4 possible cases that land along the line
:
or
. We will count the number of ways to end up at
and
, multiply them by 4 to account for the other quadrants, and add this to the number of ways to end up at
.
- Case 1: The token ends at
. In order for the token to end up here, it must have had 3 right moves, and 3 up moves. In other words, the total number of ways to get here is the ways to rearrange the letters in the sequence
, which is
- Case 2: The token ends at
. In order for the token to end up here, it could have had 2 up moves, 3 right moves, and 1 left move; or 2 right moves, 3 up moves, and 1 down move. Thus, the total number of ways to get here is sum of the ways to rearrange the letters in the sequences
and
, both of which are
, for a total of
possibilities.
- Case 3: The token ends at
. In order for the token to end up here, it could have had:
- 1 right move, 3 up moves, and 2 down moves. In this case, the number of ways to rearrange the letters in the sequence
is
- 1 up move, 3 right moves, and 2 left moves. In this case, the number of ways to rearrange the letters in the sequence
is
- 2 right moves, 1 left move, 2 up moves, and 1 down move. In this case, the number of ways to rearrange the letters in the sequence
is
- 1 right move, 3 up moves, and 2 down moves. In this case, the number of ways to rearrange the letters in the sequence
Thus, the total number of ways to end up at is
.
- Case 4: The token ends at
. In order for the token to end up here, it could have had:
- 3 right moves and 3 left moves. In this case, the number of ways to rearrange the letters in the sequence
is
- 3 up moves and 3 down moves. In this case, the number of ways to rearrange the letters in the sequence
is
- 1 right move, 1 left move, 2 up moves, and 2 down moves. In this case, the number of ways to rearrange the letters in the sequence
is
- 1 up move, 1 down move, 2 right moves, and 2 left moves. In this case, the number of ways to rearrange the letters in the sequence
is
- 3 right moves and 3 left moves. In this case, the number of ways to rearrange the letters in the sequence
Thus, the total number of ways to end up at is
.
Adding these cases together, we get that the total number of ways to end up on is
. Thus, our probability is
. When this fraction is fully reduced, it is
, so our answer is
Solution (casework 2)
Make a chart:
R (+) L (-) | U (+) D (-) +1 +1 +1 | +1 +1 + 1 +1 +1 -1 | +1 +1 -1 +1 -1 -1 | +1 -1 -1 -1 -1 -1 | -1 -1 -1
+1 +1 +1 -1 | +1 +1 +1 +1 -1 -1 | +1 -1 +1 -1 -1 -1 | -1 -1 (x2 symmetry by swapping R-L and U-D)
+1 +1 +1 -1 -1 | +1 +1 +1 -1 -1 -1 | -1 (x2 symmetry)
+1 +1 +1 -1 -1 -1 | (x2 symmetry)
Note we have to multiply by two to account for cases like +1 +1 +1 | -1 -1 -1.
Now, we use multinomials and do the arithmetic:
Thus, there are 800 + 840 + 480 + 40 = 2160 cases. Because there are total cases, the answer is
, so the answer is
See also
2014 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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.