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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| + | ==Comments(by PrincessBelle)== | ||
| + | I agree! I used Solution <math>2</math> to figure out this problem! | ||
| + | |||
| + | ~PrincessBelle | ||
==Video Solution by WhyMath== | ==Video Solution by WhyMath== | ||
Revision as of 13:54, 9 June 2021
Contents
Problem
How many integers between 
and 
have four distinct digits arranged in increasing order? (For example, 
is one integer.)
Solution 1
Firstly, observe that the second digit of such a number cannot be 
or 
, because the digits must be distinct and increasing. The second digit also cannot be 
as the number must be less than , so it must be 
. It remains to choose the latter two digits, which must be distinct digits from 
. That can be done in 
ways; there is then only way to order the digits, namely in increasing order. This means the answer is 
.
Solution 2 (without using the "choose" function)
As in Solution 1, we find that the first two digits must be 
, and the third digit must be at least . If it is , then there are 
choices for the last digit, namely , 
, 
, 
, or 
. Similarly, if the third digit is , there are choices for the last digit, namely , , , and ; if , there are choices; if , there are choices; and if , there is choice. It follows that the total number of such integers is 
.
Comments(by PrincessBelle)
I agree! I used Solution to figure out this problem!
~PrincessBelle
Video Solution by WhyMath
~savannahsolver
Video Solution
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
