# 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 $1$ of 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 $3$. $${7\choose 4} {2\choose 1}= 35\times2.$$

Thus, $\frac{35\times2}{210} = \frac{1}{3}$, which is $\fbox{C}$.

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