2003 AMC 12A Problems/Problem 22
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[hide]Problem
Objects and move simultaneously in the coordinate plane via a sequence of steps, each of length one. Object starts at and each of its steps is either right or up, both equally likely. Object starts at and each of its steps is either to the left or down, both equally likely. Which of the following is closest to the probability that the objects meet?
Solution 1
If and meet, their paths connect and There are such paths. Since the path is units long, they must meet after each travels units, so the probability is .
Note: The number of paths, comes from the fact that there must be 5 ups/downs and 7 lefts/rights in one path. WLOG, for Object A, the number of paths would be the amount of combinations of the sequence of letters with 5 "U"s 7 "R"s (i.e. UUUUURRRRRRR). This is , which is equivalent to . ~bearjere
Solution 2 (Generating Functions)
We know that the sum of the vertical steps must be equal to . We also know that they must take steps each. Since moving vertically or horizontally is equally likely, we can write all the possible paths as a generating function:
Where we need to extract the coefficient. By the binomial coefficient theorem, that term is paths. Since there are also paths, we have:
Solution 3 (Alcumus)
Since there are twelve steps between and , and can meet only after they have each moved six steps. The possible meeting places are , , , , , and . Let and denote the number of paths to from and , respectively. Since has to take steps to the right and has to take steps down, the number of ways in which and can meet at isSince and can each take paths in six steps, the probability that they meet is
See Also
2003 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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 |
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