# 2019 AIME I Problems/Problem 2

## Problem 2

Jenn randomly chooses a number $J$ from $1, 2, 3,\ldots, 19, 20$. Bela then randomly chooses a number $B$ from $1, 2, 3,\ldots, 19, 20$ distinct from $J$. The value of $B - J$ is at least $2$ with a probability that can be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

## Solution

Realize that by symmetry, the desired probability is equal to the probability that $J - B$ is at most $-2$, which is $\frac{1-P}{2}$ where $P$ is the probability that $B$ and $J$ differ by 1 (no zero, because the two numbers are distinct). There are $20 * 19 = 380$ total possible combinations of $B$ and $J$, and $1 + 18 * 2 + 1 = 38$ ones that form $P$, so $P = \frac{38}{380} = \frac{1}{10}$. Therefore the answer is $\frac{9}{20} \rightarrow \boxed{029}$.

## Solution 2

This problem is basically asking how many ways there are to choose 2 distinct elements from a 20 element set such that no 2 elements are adjacent. Using the well-known formula $\dbinom{n-k+1}{k}$, there are $\dbinom{20-2+1}{2} = \dbinom{19}{2} = 171$ ways. Dividing 171 by 380, our desired probability is $\frac{171}{380} = \frac{9}{20}$. Thus, our answer is $9+20=29$. -Fidgetboss_4000

## Solution 3

Simply create a grid using graph paper, with 20 columns for J and 20 rows for B. Since we know that $B != C$, we cross out the diagonal line from the first column of the first row to the twentieth column of the last row. Now, since $B - J$ must be at least $2$, we can mark the line where $B - J = 2$. Now we sum the number of squares that are on this line and below it. We get $171$. Then we find the number of total squares, which is $400 - 20 = 380$. Finally, we compute $\frac{171}{380}$, which simplifies to $\frac{9}{20}$. Our answer is $9+20=29$.