Difference between revisions of "1997 AJHSME Problems/Problem 9"

Latest revision as of 17:47, 28 November 2024

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 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:

$ABC$

$ACB$

$BAC$

$BCA$

$CAB$

$CBA$

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

Solution 3

We look at the first place in the line, there is only one correct person who belongs in this place, but without the restriction there can be three different people who can go in this spot. So we have a $\frac{1}{3}$ chance of getting the first place right. Now look at the second place, there is only one correct person who belongs in this place, but without the restriction there can be two different people who can go in this place, as one person is already in the first place. So we have a $\frac{1}{2}$ chance of getting the second place right. Now look at the third place, there is only one correct person who belongs in this place, but since all the other spots are filled there is only one person who can be in this place at all. So we have a $\frac{1}{1}$ chance of getting the third place right. Since the total probability of getting the order depends on the probability of each place having the right person we should multiply these three fractions. $\frac{1}{3}\times\frac{1}{2}\times\frac{1}{1}$ is the probability of getting the order right. Simplifying this we get that the answer is $\frac{1}{6}$ or $\boxed{C}$ . ~blankbox

@@ Line 27: / Line 27: @@
 Only <math>1</math> of those <math>6</math> lists is in alphabetical order, giving an answer of <math>\frac{1}{6}</math> or <math>\boxed{C}</math>
+==Solution 3==
+We look at the first place in the line, there is only one correct person who belongs in this place, but without the restriction there can be three different people who can go in this spot. So we have a <math>\frac{1}{3}</math> chance of getting the first place right.
+Now look at the second place, there is only one correct person who belongs in this place, but without the restriction there can be two different people who can go in this place, as one person is already in the first place. So we have a <math>\frac{1}{2}</math> chance of getting the second place right.
+Now look at the third place, there is only one correct person who belongs in this place, but since all the other spots are filled there is only one person who can be in this place at all. So we have a <math>\frac{1}{1}</math> chance of getting the third place right.
+Since the total probability of getting the order depends on the probability of each place having the right person we should multiply these three fractions.
+<math>\frac{1}{3}\times\frac{1}{2}\times\frac{1}{1}</math> is the probability of getting the order right.
+Simplifying this we get that the answer is <math>\frac{1}{6}</math> or <math>\boxed{C}</math>.
+~blankbox
 == See also ==

1997 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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