Using the digits $1$, $2$, $3$, $7$, $8$, $9$, and $0$, how many $4$-digit numbers can be created that are greater than $3718$?
My answer is : $(360)(1)(12)(40)= 413$
In the event that digits are not allowed to be repeated:
Let us break this into cases: - Case 1: The number is greater than or equal to $4000$
Step 1: Pick the first digit. Options are $7,8,9$ for a total of $3$ options
Step 2: Pick the second digit. Options are any of $1,2,3,7,8,9,0$ except what was picked in step 1 for a total of $6$ options.
Step 3: Pick the third digit. Options are any of the available digits except those already picked for a total of $5$ options.
Step 4: Pick the fourth digit. Options are any of the available digits except those already picked for a total of $4$ options.
This gives a total of $3\cdot 6\cdot 5\cdot 4$ numbers in this case
Case 2: The number is less than $4000$
Case 2a: The number starts with $371\square$. There is only one possibility: $3719$
Case 2b: The number starts with $37\square\square$ and the third digit is larger than $1$. There are $3\cdot 4$ possibilities here
Case 2c: The number starts with $3\square\square\square$ and the second digit is larger than $7$. There are $2\cdot 5\cdot 4$ possibilities.
This gives a final grand total of $3\cdot 6\cdot 5\cdot 4 + 1 + 3\cdot 4 + 2\cdot 5\cdot 4=413$ numbers.
Your mistake was that the number you counted was instead the number of numbers greater than $3000$ that use the available digits, but not the number of numbers greater than $3718$. There are numbers that you counted between $3000$ and $3718$ that you shouldn't have.