Question

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 $a + b?$

$5$

$6$

$11$

$16$

$19$

Solution:

The set $S$ contains $4 \cdot 4! = 96$ permutations, since the first term has $4$ choices and the remaining four terms can be arranged in $4!$ ways.

For the second term to be $2,$ the first term must be $3, 4,$ or $5$ (not $1,$ not $2$ ), giving $3$ choices, and the remaining three terms can be arranged in $3!$ ways: $3 \cdot 3! = 18.$

The probability is $\dfrac{18}{96} = \dfrac{3}{16},$ so $a + b = 3 + 16 = 19.$

Thus, the correct answer is E.

2003 AMC 12B Problem 19

Solution:

Problem 19 in Other Years