Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2 (without using the "choose" function))
(→‎Categorized problem)
 
(15 intermediate revisions by 10 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
−
How many integers between <math>2020</math> and <math>2400</math> have four distinct digits arranged in increasing order? (For example, <math>2347</math> is one integer.)
+
How many integers between <math>2020</math> and <math>2400</math> have four distinct digits arranged in increasing order? (For example, <math>2347</math> is one integer.).
  
 
<math>\textbf{(A) }\text{9} \qquad \textbf{(B) }\text{10} \qquad \textbf{(C) }\text{15} \qquad \textbf{(D) }\text{21}\qquad \textbf{(E) }\text{28}</math>
 
<math>\textbf{(A) }\text{9} \qquad \textbf{(B) }\text{10} \qquad \textbf{(C) }\text{15} \qquad \textbf{(D) }\text{21}\qquad \textbf{(E) }\text{28}</math>
  
 
==Solution 1==
 
==Solution 1==
−
Firstly, observe that the second digit of such a number cannot be <math>1</math> or <math>2</math>, because the digits must be distinct and increasing. The second digit also cannot be <math>4</math> as the number must be less than <math>2400</math>, so it must be <math>3</math>. It remains to choose the latter two digits, which must be <math>2</math> distinct digits from <math>\left\{4,5,6,7,8,9\right\}</math>. That can be done in <math>\binom{6}{2} = \frac{6 \cdot 5}{2 \cdot 1} = 15</math> ways; there is then only <math>1</math> way to order the digits, namely in increasing order. This means the answer is <math>\boxed{\textbf{(C) }15}</math>.
+
Firstly, we can observe that the second digit of such a number cannot be <math>1</math> or <math>2</math> because the digits must be distinct and in increasing order. The second digit also cannot be <math>4</math> as the number must be less than <math>2400</math>, so the second digit must be <math>3</math>. It remains to choose the latter two digits, which must be <math>2</math> distinct digits from <math>\left\{4,5,6,7,8,9\right\}</math>. That can be done in <math>\binom{6}{2} = \frac{6 \cdot 5}{2 \cdot 1} = 15</math> ways; there is then only <math>1</math> way to order the digits, namely in increasing order. This means the answer is <math>\boxed{\textbf{(C) }15}</math>.
  
 
==Solution 2 (without using the "choose" function)==
 
==Solution 2 (without using the "choose" function)==
−
As in Solution 1, we find that the first two digits must be <math>23</math>, and the third digit must be at least <math>4</math>. If it is <math>4</math>, then there are <math>5</math> choices for the last digit, namely <math>5</math>, <math>6</math>, <math>7</math>, <math>8</math>, or <math>9</math>. Similarly, if the third digit is <math>5</math>, there are <math>4</math> choices for the last digit, namely <math>6</math>, <math>7</math>, <math>8</math>, and <math>9</math>; if <math>6</math>, there are <math>3</math> choices; if <math>7</math>, there are <math>2</math> choices; and if <math>8</math>, there is <math>1</math> choice. It follows that the total number of such integers is <math>5+4+3+2+1=\boxed{\textbf{(C) }15}</math>.
+
As in Solution 1, we find that the first two digits must be <math>23</math>, and the third digit must be at least <math>4</math> because the digits can not repeat. If it is <math>4</math>, then there are <math>5</math> choices for the last digit, namely <math>5</math>, <math>6</math>, <math>7</math>, <math>8</math>, or <math>9</math>. Similarly, if the third digit is <math>5</math>, there are <math>4</math> choices for the last digit, namely <math>6</math>, <math>7</math>, <math>8</math>, and <math>9</math>; if <math>6</math>, there are <math>3</math> choices; if <math>7</math>, there are <math>2</math> choices; and if <math>8</math>, there is <math>1</math> choice. It follows that the total number of such integers is <math>5+4+3+2+1=\boxed{\textbf{(C) }15}</math>.
  
−
==Solution 3==
+
==Video Solution by NiuniuMaths (Easy to understand!)==
−
Distinct Digits means the numbers are in increasing order, but it cannot repeat the same note. For example, 1234 and 1235 can be a distinct digit, while 1223 cannot. So, the answers are 2345, 2346, 2347, 2348, 2349, 2356, 2357, 2358, 2359, 2367, 2368, 2369, 2378, 2379, and 2389. This means 15 distinct digits from 2020 to 2400. Therefore, the answer is <math>\boxed{\textbf{(C) }15}</math>.
+
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==
 
==Video Solution by WhyMath==
Line 25: Line 37:
  
 
~Interstigation
 
~Interstigation
 +
 +
==Video Solution by STEMbreezy==
 +
https://youtu.be/U27z1hwMXKY?list=PLFcinOE4FNL0TkI-_yKVEYyA_QCS9mBNS&t=85
 +
 +
~STEMbreezy
  
 
==See also==  
 
==See also==  
 
{{AMC8 box|year=2020|num-b=6|num-a=8}}
 
{{AMC8 box|year=2020|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
 +
[[Category:Introductory Combinatorics Problems]]

Latest revision as of 19:43, 4 June 2025

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, we can observe that the second digit of such a number cannot be $1$ or $2$ because the digits must be distinct and in increasing order. The second digit also cannot be $4$ as the number must be less than $2400$, so the second digit 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 $4$ because the digits can not repeat. 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

See also

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_7&oldid=249727"