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

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==Problem 7==
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==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.)
  
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==Solution 1==
 
==Solution 1==
First, observe that the second digit of the four digit number cannot be a <math>1</math> or a <math>2</math> because the digits must be distinct and increasing. The second digit also cannot be a <math>4</math> because the number must be less than <math>2400</math>. Thus, the second digit must be <math>3</math>. If we place a <math>4</math> in the third digit then there are 5 ways to select the last digit, namely the last digit could then be <math>5,6,7,8,</math> or <math>9</math>. If we place a <math>5</math> in the third digit, there are 4 ways to select the last digit, namely the last digit could then be <math>6,7,8,</math> or <math>9</math>. Similarly, if the third digit is <math>6</math>, there are 3 ways to select the last digit, etc. Thus, it follows that the total number of valid numbers is <math>5+4+3+2+1=15\implies\boxed{\textbf{(C) }15}</math>.<br>
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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>.
~[http://artofproblemsolving.com/community/user/jmansuri junaidmansuri]
 
  
==Solution 2==
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==Solution 2 (without using the "choose" function)==
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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>.
  
Notice that the number is of the form <math>23AB</math> were <math>A>B>3</math>. We have <math>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]</math>. Counting the numbers in the brackets, the answer is <math>5+4+3+2+1=\textbf{(C) }15</math>.
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==Video Solution by WhyMath==
-oceanxia
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https://youtu.be/FjmBtgrGfCs
-franzliszt
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~savannahsolver
  
 
==Video Solution==
 
==Video Solution==
https://youtu.be/FjmBtgrGfCs
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https://youtu.be/61c1MR9tne8 ~ The Learning Royal
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==Video Solution by Interstigation==
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https://youtu.be/YnwkBZTv5Fw?t=251
  
~savannahsolver
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~Interstigation
  
 
==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}}
 
 
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.
 
 
 
~itsmemasterS
 

Latest revision as of 18:41, 15 January 2022

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 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

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

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