Answer:
a) 4536
b) 2016
c) 3528
Step-by-step explanation:
For these types of problems, I recommend drawing underscores equivalent to the # of digits of the number, and then below the underscore writing the total number of possibilities that can take that position. It helps me visualize a lot better
a) we want the total number of possible arrangements.
The first spot can contain any number 1-9 (because 0 cant be the first number), so there are 9 possibilities.
The second spot can contain any number 0-9 minus whatever the first number was, so there are again 9 possibilities.
The third spot can contain any number 0-9 minus the two numbers that were used previously, so there are 8 possibilities.
The fourth spot can contain any number 0-9 minus the three numbers that were used previously, so there are 7 possibilities.
The total number of possibilities would then be [tex]9*9*8*7 = 4536[/tex].
b) Once again the process for the 2nd-4th digits are the same as the previous problem, so i will only discuss the first digit.
The first digit can contain 6,7,8 and 9, meaning there are 4 possibilities. Because there is no repetition, any 4 digit number that starts with 6,7,8, or 9 in this scenario will always be greater than 6000.
Thus, the total number of possibilities would be [tex]4*9*8*7 = 2016[/tex].
c) Once again, i will only discuss the first digit. 2nd-4th digits are same as previous 2 problems for the same reasons.
To have a number below 8000, the numbers can only start from 0-7. However, because we cannot start with 0, we are limited to 1-7. That means there are a total of 7 possibilities for the first digit.
Thus, the total number of possibilities would be [tex]7*9*8*7 = 3528[/tex]
Hoped this helped
