2003 AMC 12B Problems/Problem 19

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 $3 + 16 = 19 \Rightarrow \mathrm{(E)}$

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

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2003 AMC 12B Problems/Problem 19

Contents

Problem

Solution

Solution 2

See also