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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 