2020 AMC 8 Problems/Problem 7

1 Problem
2 Solution 1
3 Solution 2 (without using the "choose" function)
4 Video Solution by NiuniuMaths (Easy to understand!)
5 Video Solution by Math-X (First understand the problem!!!)
6 Video Solution (💩Fast💩)
7 Video Solution by WhyMath
8 Video Solution
9 Video Solution by Interstigation
10 Video Solution by STEMbreezy
11 See also

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

Video Solution by NiuniuMaths (Easy to understand!)

https://www.youtube.com/watch?v=8hgK6rESdek&t=9s

~NiuniuMaths

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/UnVo6jZ3Wnk?si=xBEcgPkL367f3Zp8&t=713

~Math-X

Video Solution (💩Fast💩)

https://youtu.be/QqBpLTQojHg

~Education, the Study of Everything

Video Solution by WhyMath

https://youtu.be/FjmBtgrGfCs

~savannahsolver

Video Solution

https://youtu.be/61c1MR9tne8 ~ The Learning Royal

Video Solution by Interstigation

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

~Interstigation

Video Solution by STEMbreezy

https://youtu.be/U27z1hwMXKY?list=PLFcinOE4FNL0TkI-_yKVEYyA_QCS9mBNS&t=85

~STEMbreezy

2020 AMC 8 (Problems • Answer Key • Resources)
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

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