2003 AMC 12B Problems/Problem 19

Revision as of 22:11, 5 February 2008 by Azjps (Talk | contribs)

(diff) ← Older revision | Current revision (diff) | Newer revision → (diff)

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 $a+b$ ?

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 $a+b=19 \Rightarrow \mathrm{(E)}$ .