2020 AMC 8 Problems/Problem 7
Contents
Problem 7
How many integers between 
and 
have four distinct digits arranged in increasing order? (For example, 
is one integer.)
Solution 1
First, observe that the second digit of the four digit number cannot be a 
or a 
because the digits must be distinct and increasing. The second digit also cannot be a 
because the number must be less than . Thus, the second digit must be 
. If we place a in the third digit then there are 5 ways to select the last digit, namely the last digit could then be 
or 
. If we place a 
in the third digit, there are 4 ways to select the last digit, namely the last digit could then be 
or . Similarly, if the third digit is 
, there are 3 ways to select the last digit, etc. Thus, it follows that the total number of valid numbers is 
.
~ junaidmansuri
Solution 2
Notice that the number is of the form 
were 
. We have ![$A=4,B\in [5,9];A=5,B\in [6,9];A=6,B\in [7,9];A=7,B\in [8,9];A=8,B\in [9]$](http://latex.artofproblemsolving.com/3/7/7/3774944bae3612163c312ad8c29a2a881ece4de7.png)
. Counting the numbers in the brackets, the answer is 
.
-franzliszt
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. 
The thousands place (first digit) has to be a 2 (2020-2400).
Since the thousands digit is 2, the next digit must be a 3 (not 4 or onwards because that will go over the range given).
The next digit has to be from 4, 5, 6, 7, or 8. For each of the cases, you get a total of 15 possibilities, which gives you the answer C.
~itsmemasterS
