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

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==Problem==
 
==Problem==
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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.)
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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>
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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>. 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>.
  
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==Comments(by PrincessBelle)==
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==Video Solution by NiuniuMaths (Easy to understand!)==
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I agree! I used Solution <math>2</math> to figure out this problem!
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https://www.youtube.com/watch?v=8hgK6rESdek&t=9s
  
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~PrincessBelle
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~NiuniuMaths
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 +
==Video Solution by Math-X (First understand the problem!!!)==
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https://youtu.be/UnVo6jZ3Wnk?si=xBEcgPkL367f3Zp8&t=713
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 +
~Math-X
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 +
==Video Solution (🚀Fast🚀)==
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https://youtu.be/QqBpLTQojHg
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 +
~Education, the Study of Everything
  
 
==Video Solution by WhyMath==
 
==Video Solution by WhyMath==
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==Video Solution==
 
==Video Solution==
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https://youtu.be/61c1MR9tne8
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https://youtu.be/61c1MR9tne8 ~ The Learning Royal
  
 
==Video Solution by Interstigation==
 
==Video Solution by Interstigation==
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~Interstigation
 
~Interstigation
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==Video Solution by STEMbreezy==
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https://youtu.be/U27z1hwMXKY?list=PLFcinOE4FNL0TkI-_yKVEYyA_QCS9mBNS&t=85
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 +
~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}}

Latest revision as of 19:02, 12 June 2024

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

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

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

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