2020 AMC 8 Problems/Problem 7
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2 (without using the "choose" function)
- 4 Video Solution by NiuniuMaths (Easy to understand!)
- 5 Video Solution by Math-X (First understand the problem!!!)
- 6 Video Solution (🚀Fast🚀)
- 7 Video Solution by WhyMath
- 8 Video Solution
- 9 Video Solution by Interstigation
- 10 Video Solution by STEMbreezy
- 11 See also
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 .
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🚀)
~Education, the Study of Everything
Video Solution by WhyMath
~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.