Difference between revisions of "2003 AMC 12A Problems/Problem 22"
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| − | == Solution == | + | == Solution 1== |
| − | If <math>A</math> and <math>B</math> meet, their paths connect <math>(0,0)</math> and <math>(5,7).</math> There are <math>\binom{12}{5}=792</math> such paths, so the probability is <math>\frac{792}{2^{6}\cdot 2^{6}} \approx 0.20 \Rightarrow \boxed{C}</math> | + | If <math>A</math> and <math>B</math> meet, their paths connect <math>(0,0)</math> and <math>(5,7).</math> There are <math>\binom{12}{5}=792</math> such paths. Since the path is <math>12</math> units long, they must meet after each travels <math>6</math> units, so the probability is <math>\frac{792}{2^{6}\cdot 2^{6}} \approx 0.20 \Rightarrow \boxed{C}</math>. |
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| + | Note: The number of paths, <math>\binom{12}{5}</math> 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 <math>\frac{12!}{5!7!}</math>, which is equivalent to <math>\binom{12}{5}</math>. | ||
| + | ~bearjere | ||
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| + | == Solution 2 (Generating Functions) == | ||
| + | |||
| + | We know that the sum of the vertical steps must be equal to <math>7</math>. We also know that they must take <math>6</math> steps each. Since moving vertically or horizontally is equally likely, we can write all the possible paths as a generating function: | ||
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| + | <cmath>P(x)=(x+1)^{12}</cmath> | ||
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| + | Where we need to extract the <math>x^5</math> coefficient. By the binomial coefficient theorem, that term is <math>\binom{12}{5}=792</math> paths. Since there are also <math>2^{12}</math> paths, we have: | ||
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| + | <math>\frac{792}{2^{12}}=\frac{792}{4096}\approx\frac{800}{4000}\approx\boxed{\text{(C) } 0.20}</math> | ||
== See Also == | == See Also == | ||
Latest revision as of 07:40, 30 June 2023
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:
![\[P(x)=(x+1)^{12}\]](http://latex.artofproblemsolving.com/f/4/e/f4e1890bbbdf5eed9c6a746dedbf5229461b53e6.png)
Where we need to extract the 
coefficient. By the binomial coefficient theorem, that term is paths. Since there are also 
paths, we have:
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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