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2020 AMC 8 Problems/Problem 7

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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 $5+4+3+2+1=15\implies\boxed{\textbf{(C) }15}$.

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 $5+4+3+2+1=\textbf{(C) }15$.


Video Solution



See also

2020 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

The thousands place (first digit) has to be a 2 (2020-2400). Since the thousands digit is 2, the next digit must be a 3 (not 4 or onwards because that will go over the range given).

The next digit has to be from 4, 5, 6, 7, or 8. For each of the cases, you get a total of 15 possibilities, which gives you the answer C.


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