# How many 4 digit numbers are there having at least one digit that is a 8 or a 9?

Table of Contents

- 1 How many 4 digit numbers are there having at least one digit that is a 8 or a 9?
- 2 How many 4 digit numbers are there such that digits are repeated at least once?
- 3 How many 4 digit numbers are there that have at least one digit that is 4 or 5?
- 4 How many at least one digit equals 8?
- 5 How do you make 4*(9^3) numbers?
- 6 How many integers are there with at least one duplicated digit?

## How many 4 digit numbers are there having at least one digit that is a 8 or a 9?

4 * 3 * 2 * 7 = 168 – This represent all numbers 0–9999 that contain exactly ONE 7 ONE 8 and ONE 9.

## How many 4 digit numbers are there such that digits are repeated at least once?

So there are 4464 integers from 1000–9999 with at least one duplicated digit. 1000 1001 1002 1003 1004 1005 1006 1007 1008 1009 1010 1011 1012 1013 1014 1015 1016 1017 1018 1019 1020 1021 1022 1030 1031 1033 1040 1041 1044 1050 1051 1055 1060 1061 1066 1070 1071 1077 1080 1081 …..

**How many four-digit numbers are there whose at least one digit is even?**

Thus, the number of odd digited 4 digit numbers would be 5*5*5*5 that is 625 ways. Now, with the even number, we have a problem. The first digit cannot be 0 other wise, the number would not be a 4 digit number. So, p can only contain 4 digits (2,4,6,8).

**How many 4 digit numbers contain at least one 7?**

The answer is that there are 198 four-digit integers that contain at least one 7, one 8, and one 9. The answer is that there are 198 four-digit integers that contain at least one 7, one 8, and one 9.

### How many 4 digit numbers are there that have at least one digit that is 4 or 5?

How many numbers of 4 digits contains at least one “5”? My suggestion is: First I count all the numbers of type 5xxx (where x can be a number between 0 and 9). There is 999.

### How many at least one digit equals 8?

For thousand’s, hundred’s, ten’s and unit’s places there are 9 choices each (only ‘8′ excluded). Thus, there are (8 * 9 * 9 * 9 * 9) = 52488 five digit numbers which have no digit ‘8′ in them. Therefore, there are (90000 – 52488) = 37512 five digit numbers which have at least one digit ‘8′ in them.

**How many 4 digit codes can be formed from the digits 1/7 If digits Cannot be repeated?**

Answer: There are 9,999 different combinations in a 4-digit lock.

**How many 4 digit numbers have a 1 in the digit?**

Now, all the 4 digit numbers with a 1 in the Tens’ digit, which is 1, 010 to 1, 019, 1, 110 to 1, 119, 1, 210 to 1, 219 ,…, 9, 910 to 9, 919. This is 90 groups of 10, aka 900 values. However, since we already counted all the 4 digit numbers with a 1 in the One Thousands’ digit, we exclude those 100 values.

#### How do you make 4*(9^3) numbers?

As repetition is not specified, you can make 4* ( 9^3) numbers because combinatorics. First three numbers can be anything because eve and odd is dependent upon the last digit. and you multiply by 4 because the last digit can be 2 4 6 8 to make it even.

#### How many integers are there with at least one duplicated digit?

We’ve put together a list of 8 money apps to get you on the path towards a bright financial future. Generate all the numbers 1000–9999, remove all numbers with no duplicate digits, then count the remaining numbers: So there are 4464 integers from 1000–9999 with at least one duplicated digit.

**How many 3 digit numbers can be formed with one digit repeated?**

We need at least one digit repeated and the easiest way to find that is to subtract the number of 4 digit numbers with no digits repeating from the total number of four digit numbers. How many three digit numbers can be formed using the digits 2, 3, 4, 5, and 6 if the digits can be repeated? The answer is 125.