2020 AMC 8 Problems/Problem 7

Problem

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

Firstly, observe that the second digit of such a number cannot be $1$ or $2$ , because the digits must be distinct and increasing. The second digit also cannot be $4$ as the number must be less than $2400$ , so it must be $3$ . It remains to choose the latter two digits, which must be $2$ distinct digits from $\left\{4,5,6,7,8,9\right\}$ . That can be done in $\binom{6}{2} = \frac{6 \cdot 5}{2 \cdot 1} = 15$ ways; there is then only $1$ way to order the digits, namely in increasing order. This means the answer is $\boxed{\textbf{(C) }15}$ .

Solution 2 (without using the "choose" function)

As in Solution 1, we find that the first two digits must be $23$ , and the third digit must be at least $4$ . If it is $4$ , then there are $5$ choices for the last digit, namely $5$ , $6$ , $7$ , $8$ , or $9$ . Similarly, if the third digit is $5$ , there are $4$ choices for the last digit, namely $6$ , $7$ , $8$ , and $9$ ; if $6$ , there are $3$ choices; if $7$ , there are $2$ choices; and if $8$ , there is $1$ choice. It follows that the total number of such integers is $5+4+3+2+1=\boxed{\textbf{(C) }15}$ .

Comments

I agree! I used Solution $2$ to figure out this problem!

~PrincessBelle

I did Solution $2$ too! And the answer is right!

~lordofeverything

The difference is that the first is faster but more advanced, while the second is simpler but slower

Video Solution by WhyMath

https://youtu.be/FjmBtgrGfCs

~savannahsolver

Video Solution

https://youtu.be/61c1MR9tne8

Video Solution by Interstigation

https://youtu.be/YnwkBZTv5Fw?t=251

~Interstigation

2020 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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All AJHSME/AMC 8 Problems and Solutions

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