# How many 3 digit numbers can be formed from the digits 4 6 and 8?

## How many 3 digit numbers can be formed from the digits 4 6 and 8?

The answer is 24. SOLUTION: See, the hundreds place can be filled up with any of the four numbers. So, 4 choices for hundreds place.

What is the sum of all the 4 digit numbers that can be formed using all of the digits?

There are total 4^4 = 256 combinations. Each digit comes in every place 4^3 = 64 times. Sum of all digits in each place = 64 ( 1+2+3+4 ) = 640.

What is the sum of all 3 digit numbers formed by using the digits 1 2 3 without repetition?

The sum of all 3 digit-numbers formed using 1, 2, 3 repetition is allowed) is 5994 (2) 5894 5694 (4) 6694.

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### What is the sum of all 3 digit numbers that leave?

Thus we got the sum of all 3 digit numbers that leave a remainder of 2 when divided by 3 is 164,850. Option (d) is the correct answer.

How many 3 digit number can be formed using the digits?

Thus, The total number of 3-digit numbers that can be formed = 5 × 4 × 3 = 60. Question 3: How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated? Let 3-digit number be ABC.

What is the sum of all 3 digit numbers that leave remainder of 2 when divided by 3?

What is the sum of all 3 digit numbers that leave a remainder of ‘2’ when divided by 3? The lowest 3 digit number is 101 and the highest number is 998. Hence, the sum of all 3 digit numbers that leave a remainder of 2 when divided by 3 is 164,850.

#### How do you find the sum of a 3 digit number in C?

Sum of digits algorithm

1. Step 1: Get number by user.
2. Step 2: Get the modulus/remainder of the number.
3. Step 3: sum the remainder of the number.
4. Step 4: Divide the number by 10.
5. Step 5: Repeat the step 2 while number is greater than 0.
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What is the sum of all numbers formed by using 4 digits?

Sol: Sum of the numbers formed by taking all the given n digits is (sum of all the n digits) x (n-1) ! x (111…..n times). Here n = 4. and Sum of 4 digits = 16 The sum of all the numbers which can be formed by using the digits 1, 3, 5, 7 is = (16) x (4 – 1)! x (1111)

What is the sum of all possible numbers using n distinct digits?

“ If all the possible n-digit numbers using n distinct digits are formed, the sum of all the numbers so formed is equal to (n-1)! x {sum of all the digits} x {111…….} n times. Example 1: What would be the sum of all the numbers which can be formed by using the digits 1, 3, 5, 7 all at a time and which have no digits repeated?

## What is the contribution of 3 4 and 5 to the sum?

In a similar manner, the contribution of the 3,4, and 5 to the sum will respectively be 3! x 3333, 3! x 4444, 3! x 5555, i.e., “ If all the possible n-digit numbers using n distinct digits are formed, the sum of all the numbers so formed is equal to (n-1)! x {sum of all the digits} x {111…….} n times.

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What is contribution to the sum by 6 at Hundred’s digit?

And, contribution to the sum by 6 at hundred’s digit: 4! x 600 = 14400 Subtracting these numbers sum from the original sum will give the answer i.e. Example 4: What would be the sum formed by using the digits 2,3,4,6,0, in which the position of 0 is fixed at the hundred’s digit.