 # Quick Answer: How Many Numbers Can Be Formed From 1 2 3 4 5 Without Repetition When The Digit At Unit’S Place Must Be Greater Than The In The Ten’S Place?

## How many 13 digit numbers are possible by using the digits 1 2 3 4 5 which are divisible by 4 if repetition of digits is allowed?

How many 13 digit numbers are possible by using the digits 1,2,3,4,5 which are divisible by 4 if repetition of digits is allowed.

hence the total no of digits allowed are (5^11)*5 = 5^12.

## How many 9 digit numbers are possible by using the digits 1 2 3 4 5 which are divisible by 4 if the repetition is allowed?

If the units digit is 4, only 24 and 44 are acceptable as the last two digits. So, total 5 cases for the last two digits. Now, the first 7 digits of the 9 digit number can be filled in 57 ways as each place can be filled by 1, 2, 3, 4 or 5.

## What is the sum of all the 4 digit numbers?

Four digit numbers =4⋅3⋅2⋅1=24 ways we can form a four digit number. Since it’s a 4 digit number, each digit will appear 6=24/4 times in each of units, tens, hundreds, and thousands place. Therefore, the sum of digits in the units place is 6(1+2+5+6)=84.

## What is the sum of all the 4 digit numbers which can formed with the digits 1 2 3 4 without repetition?

We get that the sum of all the 4-digit numbers formed using the digits 2, 3, 4, and 5 (without repetition) is 93, 324.

## How many 3 digit numbers can be formed from the digits 1 2 3 4 and 5 of the digits are unique?

That way, The number of 3 digit number that we can form by just using 1,2,3,4 is 4 * 4 * 4 = 64. 24 such numbers can be formed.

## How many 4 digit numbers can be formed with repetition?

There will be as many 4-digit numbers as there are permutations of 5 different digits taken 4 at a time. 5 5 4 5 1 P ! ! ! ! = 5 × 4 × 3 × 1 × 1 = 120 Among the 4-digit numbers formed by using the digits, 1, 2, 3, 4, 5, even numbers end with either 2 or 4.

## How many 9 digit numbers are there?

1 billion 9There are 1 billion 9 digit numbers (000,000,000 through 999,999,999). There are 45 different combinations of two different numerals (10 x 9 divided by 2). There are 512 (2 to the 9th power) different permutations for any two numbers to be used in a 9 digit number. 45 x 512 = 23,040 – 10 = 23,030.

## What are all the combinations for a 3 Number Lock?

There are about 1,000 combinations possible for a 3-digit number/code. There are 10,000 combinations possible for a 4-digit number/code.

## How many numbers can be formed with 3 digits?

Therefore, we use the formula n^ r, where n = 5 and r = 3. Plugging these values in gives: 5 ^ 3 = 5 * 5 * 5 = 125. Therefore, we can make 125 3-digit numbers from the numbers 3, 7, 0, 2, and 9.

## How many combinations of the numbers 1 2 3 4 are there?

Explanation: If we are looking at the number of numbers we can create using the numbers 1, 2, 3, and 4, we can calculate that the following way: for each digit (thousands, hundreds, tens, ones), we have 4 choices of numbers. And so we can create 4×4×4×4=44=256 numbers.