What is the sum of all 5 digit numbers formed using 1 2 3 4 and 5 exactly once?

Out of them, each 24 will have 1, 2, 3, 4 & 5 as ten thousands, thousands, hundreds, tens & unit digit. 1 + 2 + 3 + 4 + 5 = 15 so face value of each column = 15*24 = 360. Sum = 360 (10,000 + 1,000 + 100 + 10 + 1) = 39,99,960.

How many numbers can be made with the help of the digit 0 1 2 3 4 5 which are greater than 3000 when repetition of digit not allowed?

Answer: (3) 1380 Greater than 3000 we can write, 4-digit numbers, 5 digit numbers and 6 digit numbers.

What is the sum of the even 5 digit numbers that can be formed by using digits 0 1 2 3 4?

Total number of ways = 1,080 + 2,160 = 3,240.

What is the sum of five digit number?

Solution: The greatest 5-digit number is 99,999. The smallest 5-digit number is 10,000. Their sum is 99,999 + 10,000 = 1,09,999.

How many 5 digit numbers can be formed from the digits 1,2 3 4 5 using the digits without repetition how many are even and greater than 30000?

Total Number of Numbers which can be formed by numbers 1,2,3,4,5 (without repeating digitsi) = 5*4*3*2*! = 5! = 120.

How to find sum of all 4 digit numbers formed using 0?

To find sum of all 4 digit numbers formed using 0 1 2 3, we have to find the sum of all numbers at first, second, third and fourth places. Let us find the sum of numbers at the first place (thousand’s place). In the 18 numbers formed, we have each one of the digits (1, 2, 3) six times at the first place.

What is 24 as the sum of all 4 digit numbers?

24 is the sum of numbers at unit’s place. So 24 is multiplied 1. The method explained above is not only applicable to find the sum of all 4 digit numbers formed using 0 1 2 3. This same method can be applied to find sum of all 4 digit numbers formed using any four digits in which one of the digits is zero.

What is the sum of the numbers at thousand’s place?

In the 18 numbers formed, the digit “0” comes six times and each one of the digits (1, 2, 3) comes four times at the second place. Sum of the numbers at the second place (100’s place) : 36 is the sum of numbers at thousand’s place. So 36 is multiplied 1000.

What is the sum of all 96 numbers at random?

If you pick one at random, the expected value of the first digit is 1 + 2 + 3 + 4 4 = 5 2, and the expected value of each other digit must be 15 8, because the sum of all five digits is 0 + 1 + 2 + 3 + 4 = 5 2 + 4 ⋅ 15 8. By linearity of expectation, the sum of all 96 numbers is 96 ( 5 2 ⋅ 1000 + 15 8 ⋅ 111) = 259980.