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Difference between revisions of "1987 AIME Problems/Problem 13"

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This is much easier to solve: there are <math>31!</math> ways to order the first thirty-one numbers and <math>29!</math> ways to arrange them so that the largest number is in the 31st position and the second-largest is in the 20th. This gives us a desired probability of <math>\frac{29!}{31!} = \frac{1}{31\cdot 30} = \frac{1}{930}</math>, so the answer is <math>\boxed{931}</math>.
 
This is much easier to solve: there are <math>31!</math> ways to order the first thirty-one numbers and <math>29!</math> ways to arrange them so that the largest number is in the 31st position and the second-largest is in the 20th. This gives us a desired probability of <math>\frac{29!}{31!} = \frac{1}{31\cdot 30} = \frac{1}{930}</math>, so the answer is <math>\boxed{931}</math>.
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 +
===Remark===
 +
Note that you can solve the restated problem differently:
 +
 +
We see that the numerator is <math>1</math> because there is only <math>1</math> way to make the <math>31^{\text{st}}</math> term the largest and the <math>20^{\text{th}}</math> the second-largest. To calculate the denominator, we note that there are <math>31\times30</math> ways to pick the largest term and the second-largest term. Therefore, our answer is <math>\boxed{931}</math>.
 +
~Yiyj1
  
 
== See also ==
 
== See also ==

Revision as of 20:27, 31 December 2023

Problem

A given sequence $r_1, r_2, \dots, r_n$ of distinct real numbers can be put in ascending order by means of one or more "bubble passes". A bubble pass through a given sequence consists of comparing the second term with the first term, and exchanging them if and only if the second term is smaller, then comparing the third term with the second term and exchanging them if and only if the third term is smaller, and so on in order, through comparing the last term, $r_n$, with its current predecessor and exchanging them if and only if the last term is smaller.

The example below shows how the sequence 1, 9, 8, 7 is transformed into the sequence 1, 8, 7, 9 by one bubble pass. The numbers compared at each step are underlined.

$\underline{1 \quad 9} \quad 8 \quad 7$
$1 \quad {}\underline{9 \quad 8} \quad 7$
$1 \quad 8 \quad \underline{9 \quad 7}$
$1 \quad 8 \quad 7 \quad 9$

Suppose that $n = 40$, and that the terms of the initial sequence $r_1, r_2, \dots, r_{40}$ are distinct from one another and are in random order. Let $p/q$, in lowest terms, be the probability that the number that begins as $r_{20}$ will end up, after one bubble pass, in the $30^{\mbox{th}}$ place. Find $p + q$.

Solution

If any of $r_1, \ldots, r_{19}$ is larger than $r_{20}$, one of these numbers will be compared with $r_{20}$ on the 19th step of the first bubble pass and $r_{20}$ will be moved back to the 19th position. Thus, $r_{20}$ must be the largest of the first 20 terms. In addition, $r_{20}$ must be larger than $r_{21}, r_{22}, \ldots, r_{30}$ but smaller than $r_{31}$ in order that it move right to the 30th position but then not continue moving right to the 31st.

Thus, our problem can be restated: What is the probability that in a sequence of 31 distinct real numbers, the largest is in position 31 and the second-largest is in position 20 (the other 29 numbers are irrelevant)?

This is much easier to solve: there are $31!$ ways to order the first thirty-one numbers and $29!$ ways to arrange them so that the largest number is in the 31st position and the second-largest is in the 20th. This gives us a desired probability of $\frac{29!}{31!} = \frac{1}{31\cdot 30} = \frac{1}{930}$, so the answer is $\boxed{931}$.

Remark

Note that you can solve the restated problem differently:

We see that the numerator is $1$ because there is only $1$ way to make the $31^{\text{st}}$ term the largest and the $20^{\text{th}}$ the second-largest. To calculate the denominator, we note that there are $31\times30$ ways to pick the largest term and the second-largest term. Therefore, our answer is $\boxed{931}$. ~Yiyj1

See also

1987 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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