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 
.
Solution 3
Distinct Digits means the numbers are in increasing order, but it cannot repeat the same note. 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 .
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
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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