1997 AJHSME Problems/Problem 9


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 like Abby Adams, Bob Bott, and Carol Crock: $A$, $B$, 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:







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

See also

1997 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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All AJHSME/AMC 8 Problems and Solutions

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