# Difference between revisions of "2020 AMC 8 Problems/Problem 7"

## Problem 7

How many integers between $2020$ and $2400$ have four distinct digits arranged in increasing order? (For example, $2347$ is one integer.)

$\textbf{(A) }\text{9} \qquad \textbf{(B) }\text{10} \qquad \textbf{(C) }\text{15} \qquad \textbf{(D) }\text{21}\qquad \textbf{(E) }\text{28}$

## Solution 1

First, observe that the second digit of the four digit number cannot be a $1$ or a $2$ because the digits must be distinct and increasing. The second digit also cannot be a $4$ because the number must be less than $2400$. Thus, the second digit must be $3$. If we place a $4$ in the third digit then there are 5 ways to select the last digit, namely the last digit could then be $5,6,7,8,$ or $9$. If we place a $5$ in the third digit, there are 4 ways to select the last digit, namely the last digit could then be $6,7,8,$ or $9$. Similarly, if the third digit is $6$, there are 3 ways to select the last digit, etc. Thus, it follows that the total number of valid numbers is $5+4+3+2+1=15\implies\boxed{\textbf{(C) }15}$.
~junaidmansuri

## Solution 2

Notice that the number is of the form $23AB$ were $A>B>3$. We have $A=4,B\in [5,9];A=5,B\in [6,9];A=6,B\in [7,9];A=7,B\in [8,9];A=8,B\in [9]$. Counting the numbers in the brackets, the answer is $5+4+3+2+1=\textbf{(C) }15$. -oceanxia -franzliszt

## Video Solution

~savannahsolver

 2020 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 6 Followed byProblem 8 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