Difference between revisions of "2003 AMC 12A Problems/Problem 22"
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<cmath>P(x)=(x+1)^{12}</cmath> | <cmath>P(x)=(x+1)^{12}</cmath> | ||
− | 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: | + | 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: |
− | <math>\frac{792}{2^12}=\frac{792}{4096}\approx\frac{800}{4000} | + | <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 12:29, 5 November 2020
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 .
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:
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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