1987 AIME Problems/Problem 13

Revision as of 10:44, 11 February 2007 by JBL (talk | contribs) (solved)

Problem

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

Solution

If any of $r_1, \ldots, r_{19}$ is larger than $r_{20}$, $r_{20}$ 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 40 distinct real numbers, among the first 31, the largest is in position 31 and the second-largest is in position 20?

This is much easier to solve: there are $40 \choose 31$ ways to pick 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. There are also $9!$ ways to arrange the remaining nine members of the sequence. This is out of a total of $40!$ possible permutations of the sequence. This gives us a desired probability of $\frac{{40 \choose 31} \cdot 29! \cdot 9!}{40!} = \frac{1}{30\cdot 31} = \frac{1}{930}$, so the answer is $931$.

See also

1987 AIME (ProblemsAnswer KeyResources)
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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