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Difference between revisions of "2003 AMC 12A Problems/Problem 22"

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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>.
 
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>.
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~bearjere
  
 
== Solution 2 (Generating Functions) ==
 
== Solution 2 (Generating Functions) ==
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<cmath>P(x)=(x+1)^{12}</cmath>
 
<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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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}=\boxed{\text{(C) } 0.20}</math>
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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 $A$ and $B$ move simultaneously in the coordinate plane via a sequence of steps, each of length one. Object $A$ starts at $(0,0)$ and each of its steps is either right or up, both equally likely. Object $B$ starts at $(5,7)$ 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?

$\mathrm{(A)} \ 0.10 \qquad \mathrm{(B)} \ 0.15 \qquad \mathrm{(C)} \ 0.20 \qquad \mathrm{(D)} \ 0.25 \qquad \mathrm{(E)} \ 0.30 \qquad$

Solution 1

If $A$ and $B$ meet, their paths connect $(0,0)$ and $(5,7).$ There are $\binom{12}{5}=792$ such paths. Since the path is $12$ units long, they must meet after each travels $6$ units, so the probability is $\frac{792}{2^{6}\cdot 2^{6}} \approx 0.20 \Rightarrow \boxed{C}$.

Note: The number of paths, $\binom{12}{5}$ 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 $\frac{12!}{5!7!}$, which is equivalent to $\binom{12}{5}$. ~bearjere

Solution 2 (Generating Functions)

We know that the sum of the vertical steps must be equal to $7$. We also know that they must take $6$ steps each. Since moving vertically or horizontally is equally likely, we can write all the possible paths as a generating function:

\[P(x)=(x+1)^{12}\]

Where we need to extract the $x^5$ coefficient. By the binomial coefficient theorem, that term is $\binom{12}{5}=792$ paths. Since there are also $2^{12}$ paths, we have:

$\frac{792}{2^{12}}=\frac{792}{4096}\approx\frac{800}{4000}\approx\boxed{\text{(C) } 0.20}$

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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