# Difference between revisions of "1986 AHSME Problems/Problem 22"

## Problem

Six distinct integers are picked at random from $\{1,2,3,\ldots,10\}$. What is the probability that, among those selected, the second smallest is $3$?

$\textbf{(A)}\ \frac{1}{60}\qquad \textbf{(B)}\ \frac{1}{6}\qquad \textbf{(C)}\ \frac{1}{3}\qquad \textbf{(D)}\ \frac{1}{2}\qquad \textbf{(E)}\ \text{none of these}$

## Solution

The total number of ways to choose 6 numbers is ${10\choose 6} = 210$

Assume 3 is the SECOND lowest number. There are 5 numbers left to choose, 4 of which must be greater than 3 and 1 which must be less than 3. This is equivalent to choosing 4 numbers from the 7 numbers larger than 3, and 1 number from the 2 numbers less than 3. $${7\choose 4} {2\choose 1}= 35*2$$

Thus, $\frac{35*2}{210} = \frac{1}{3}$ $\fbox{C}$