2003 AMC 12A Problems/Problem 22

Revision as of 11:49, 5 November 2020 by Captainsnake (talk | contribs) (Added solution)

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 $\frac{792}{2^{6}\cdot 2^{6}} \approx 0.20 \Rightarrow \boxed{C}$.

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:

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

See Also

2003 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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All AMC 12 Problems and Solutions

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