2020 AMC 8 Problems/Problem 7
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 .
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 | |
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