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

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==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>. 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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==Solution 3==
 
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Distinct Digits means the numbers are in increasing order, but it cannot repeat the same number.  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>.
 
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~HappyGenius1124😉
 
  
 
==Video Solution by WhyMath==
 
==Video Solution by WhyMath==

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 (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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