1997 AJHSME Problems/Problem 9

Problem

Three students, with different names, line up single file. What is the probability that they are in alphabetical order from front-to-back?

$\text{(A)}\ \dfrac{1}{12} \qquad \text{(B)}\ \dfrac{1}{9} \qquad \text{(C)}\ \dfrac{1}{6} \qquad \text{(D)}\ \dfrac{1}{3} \qquad \text{(E)}\ \dfrac{2}{3}$

Solution 1

There are $3$ ways to pick the first student, $2$ ways to pick the second student, and $1$ way to pick the last student, for a total of $3! = 3\times2\times1 = 6$ ways to line the students up.

Only $1$ of those ways is alphabetical. Thus, the probability is $\frac{1}{6}$ or $\boxed{C}$

Solution 2

Give the students uncreative names: $A$ , $B$ , and $C$ , replacing the alphabetically first student's name with $A$ , and the alphabetically last student's name with $C$ . List all the ways they can line up:

$ABC$

$ACB$

$BAC$

$BCA$

$CAB$

$CBA$

Only $1$ of those $6$ lists is in alphabetical order, giving an answer of $\frac{1}{6}$ or $\boxed{C}$

1997 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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All AJHSME/AMC 8 Problems and Solutions

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1997 AJHSME Problems/Problem 9

Contents

Problem

Solution 1

Solution 2

See also