How many 4 digit numbers are greater then 7538 can be made using the digits, 0,3,5,7,8,9 without repetition?
well, just start listing them:
7539 -> 1
78.. -> 4P2
79.. -> 4P2
8... -> 4P3
9... -> 4P3
now add 'em up.
So the answer would be 73?
works for me.
To find the number of 4-digit numbers greater than 7538 that can be made using the given digits without repetition (0, 3, 5, 7, 8, 9), we need to consider the possible choices for each digit position.
Step 1: Count the choices for the thousands place.
Since the numbers need to be greater than 7538, the thousands place cannot be 7 or less. Therefore, there are 2 choices for the thousands place: 8 or 9.
Step 2: Count the choices for the hundreds place.
The hundreds place can be any digit other than those already chosen for the thousands place and the remaining digits (0, 3, 5). Therefore, there are 3 choices for the hundreds place.
Step 3: Count the choices for the tens place.
Again, the tens place can be any digit other than those already chosen for the thousands and hundreds places and the remaining digits (0, 3). So, there are 3 choices for the tens place.
Step 4: Count the choices for the units place.
Similarly, the units place can be any digit other than those already chosen for the thousands, hundreds, and tens places and the remaining digit (0). Hence, there is only 1 choice for the units place.
To find the total number of 4-digit numbers, we multiply the number of choices for each digit position: 2 choices for the thousands place × 3 choices for the hundreds place × 3 choices for the tens place × 1 choice for the units place.
Therefore, the total number of 4-digit numbers greater than 7538 that can be made using the digits 0, 3, 5, 7, 8, 9 without repetition is 2 × 3 × 3 × 1 = 18.