**How many numbers greater than 23,000 can be formed with the digits 1,2,3,4,5,6 if digits may repeat?**

I'm not sure how I would do this if digits may repeat but I understand how to do it with digits may not repeat.

My approachh to this: 1 can't be the first digit since then it will be less than 23000

Subtract nunber of permutations of 5 objects from permutations of 4 objects - 5^5 - 4^4? but that doesn't seem to be right.

Jasonkiln Dec 29, 2022

#2**0 **

Note that all 6 digit numbers work, so there are \(6 \times 6 \times 6 \times 6 \times 6 \times 6 = 46,656 \)numbers

Now, for the 5 digit numbers, there are \(6^5 = 7776\) numbers, but we need to count for the numbers that are less than 23,000.

If the first digit is a 1, there are \(6^4 = 1296\) numbers that don't work

If the first two digits are 21, there are \(6^3 = 216\) numbers that don't work

LIkewise, if the first two digits are 22, there are 216 numbers that don't work

So, there are \(46,656 + 7776 - 1296 - 216 - 216 = \color{brown}\boxed{52704}\)

BuilderBoi Dec 29, 2022