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# 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 lowest number. There are 5 numbers left to choose, each of which has to be greater than 3. This is equivalent to choosing 5 numbers from the 7 numbers larger than 3, ${7\choose 6} = 21$

Thus, $\frac{21}{210} = \frac{1}{10}$ $\fbox{E}$