Question

how many in all do we have with 4 digits

Answer 1

As I understand it, you’re asking how many 4 digit numbers exist, where the digits are the non-negative integers 0–9. Put another way, how many elements are in the set {0,1,2,3,…,9998,9999}?

To start, there are a total of ten digits (0–10) that we can use to make our numbers.

Now, assuming 0111 is the same as 111, since the 0 digit in front doesn’t matter, we have:

9x10x10x10=9,000 possible numbers.

If you ARE counting the 0 digit in front, it’s basically the same calculation:

10x10x10x10=10^4=10,000 possible numbers.

There are a lot of different ways to solve this problem, but since it’s in combinatorics, this approach might prove useful in that subject.

Answer 2

We have 9000 numbers with 4 digits in total.
1001 - 2000 = 1000 numbers
2001 - 3000 = 1000 numbers
3001-4000 = 1000 numbers
4001-5000 = 1000 numbers
5001-6000 = 1000 numbers
6001-7000 = 1000 numbers
7001-8000 = 1000 numbers
8001-9000 = 1000 numbers
9001-9999 = 999 numbers
Here we have to include 8001 & also 9000 to get 1000 numbers. It is same for all.
Now if we add this we will get 8,999.We have ro count the number 1000 also. So they are 9000 numbers

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