Difference between revisions of "2020 AMC 8 Problems/Problem 7"
m |
Sevenoptimus (talk | contribs) (Added an improved solution, removed duplicate solutions, and cleaned up the other solutions) |
||
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.) | ||
Line 5: | Line 5: | ||
==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 <math>2</math> distinct digits from <math>\left\{4,5,6,7,8,9\right\}</math>, which 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== | + | ==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>. | |
− | |||
− | |||
− | |||
==Video Solution== | ==Video Solution== | ||
https://youtu.be/FjmBtgrGfCs | https://youtu.be/FjmBtgrGfCs | ||
− | |||
− | |||
==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}} | ||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− |
Revision as of 06:36, 20 November 2020
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 distinct digits from , which 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 .
Video Solution
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.