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2003 AMC 12B Problems/Problem 19: Difference between revisions

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== Solution ==
== Solution ==
There are <math>4</math> choices for the first element of <math>S</math>, and for each of these choices there are <math>4!</math> ways to arrange the remaining elements. If the second element must be <math>2</math>, then there are only <math>3</math> choices for the first element and <math>3!</math> ways to arrange the remaining elements. Hence the answer is <math>\frac{3 \cdot 3!}{4 \cdot 4!} = \frac {18}{96} = \frac{3}{16}</math>, and <math>a+b=19 \Rightarrow \mathrm{(E)}</math>.
There are <math>4</math> choices for the first element of <math>S</math>, and for each of these choices there are <math>4!</math> ways to arrange the remaining elements. If the second element must be <math>2</math>, then there are only <math>3</math> choices for the first element and <math>3!</math> ways to arrange the remaining elements. Hence the answer is <math>\frac{3 \cdot 3!}{4 \cdot 4!} = \frac {18}{96} = \frac{3}{16}</math>, and <math>a+b=19 \Rightarrow \mathrm{(E)}</math>.
== Solution 2 ==
There is a <math>\frac {1}{4}</math> chance that the number <math>1</math> is the second term. Let <math>x</math> be the chance that <math>2</math> will be the second term. Since <math>3, 4,</math> and <math>5</math> are in similar situations as <math>2</math>, this becomes <math>\frac {1}{4} + 4x = 1</math>
Solving for <math>x</math>, we find it equals <math>\frac {3}{16}</math>, therefore <math>3 + 16 = 19 \Rightarrow \mathrm{(E)}</math>


== See also ==
== See also ==

Revision as of 12:19, 3 November 2013

Problem

Let $S$ be the set of permutations of the sequence $1,2,3,4,5$ for which the first term is not $1$. A permutation is chosen randomly from $S$. The probability that the second term is $2$, in lowest terms, is $a/b$. What is $a+b$?

$\mathrm{(A)}\ 5 \qquad\mathrm{(B)}\ 6 \qquad\mathrm{(C)}\ 11 \qquad\mathrm{(D)}\ 16 \qquad\mathrm{(E)}\ 19$

Solution

There are $4$ choices for the first element of $S$, and for each of these choices there are $4!$ ways to arrange the remaining elements. If the second element must be $2$, then there are only $3$ choices for the first element and $3!$ ways to arrange the remaining elements. Hence the answer is $\frac{3 \cdot 3!}{4 \cdot 4!} = \frac {18}{96} = \frac{3}{16}$, and $a+b=19 \Rightarrow \mathrm{(E)}$.


Solution 2

There is a $\frac {1}{4}$ chance that the number $1$ is the second term. Let $x$ be the chance that $2$ will be the second term. Since $3, 4,$ and $5$ are in similar situations as $2$, this becomes $\frac {1}{4} + 4x = 1$

Solving for $x$, we find it equals $\frac {3}{16}$, therefore $3 + 16 = 19 \Rightarrow \mathrm{(E)}$

See also

2003 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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