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The sum of 4 digit number containing the digits 2,4,6,8 without repetition?
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The sum of 4 digit number containing the digits 2,4,6,8 without repeti...
Sum of 4-Digit Numbers Containing the Digits 2, 4, 6, 8 without Repetition
The problem requires us to find the sum of all the 4-digit numbers that can be formed using the digits 2, 4, 6, and 8 without repetition. We can solve this problem by breaking it down into smaller parts and using some basic principles of permutations and combinations.
Step 1: Finding the Total Number of 4-Digit Numbers
- The total number of 4-digit numbers that can be formed using the digits 2, 4, 6, and 8 without repetition is equal to the number of ways we can choose 4 digits out of the 4 available digits.
- This can be found using the formula for combinations, which is nCr = n! / (r! * (n - r)!), where n is the total number of items, r is the number of items we want to choose, and ! denotes factorial.
- So, in this case, we have n = 4 and r = 4, which gives us 4! / (4! * (4 - 4)!) = 1.
- Therefore, there is only 1 way to choose 4 digits out of the 4 available digits.
Step 2: Finding the Sum of the Thousands Place
- Since we are looking for the sum of all 4-digit numbers, we need to find the sum of the digits in the thousands place.
- For a number to be a 4-digit number, it must have a digit in the thousands place.
- Since we have 4 available digits, each digit will appear in the thousands place exactly once.
- Therefore, the sum of the digits in the thousands place is equal to the sum of the available digits, which is 2 + 4 + 6 + 8 = 20.
Step 3: Finding the Sum of the Other Places
- Now, we need to find the sum of the digits in the hundreds, tens, and ones places.
- Since each digit can appear in each place only once, the sum of the digits in the hundreds, tens, and ones places will be the same.
- Therefore, we can find the sum of the digits in any one of these places and multiply it by 3 to get the total sum.
- To find the sum of the digits in any one of these places, we need to consider all the possible combinations of the remaining 3 digits.
- Since we have already used one digit in the thousands place, we have 3 digits left to choose from for the other places.
- Therefore, the total number of combinations for each place is 3! = 6.
- So, the sum of the digits in any one of these places is equal to (2 + 4 + 6 + 8) * 6 = 120.
- Therefore, the total sum of the digits in the hundreds, tens, and
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The sum of 4 digit number containing the digits 2,4,6,8 without repetition?
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The sum of 4 digit number containing the digits 2,4,6,8 without repetition? for CA Foundation 2024 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about The sum of 4 digit number containing the digits 2,4,6,8 without repetition? covers all topics & solutions for CA Foundation 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The sum of 4 digit number containing the digits 2,4,6,8 without repetition?.
The sum of 4 digit number containing the digits 2,4,6,8 without repetition? for CA Foundation 2024 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about The sum of 4 digit number containing the digits 2,4,6,8 without repetition? covers all topics & solutions for CA Foundation 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The sum of 4 digit number containing the digits 2,4,6,8 without repetition?.
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