Sum of all 4-digit numbers containing the digits 2, 4, 6, 8 without repetition:

In order to find the sum of all 4-digit numbers containing the digits 2, 4, 6, and 8 without repetition, we need to consider the possible combinations of these digits and calculate their sum.

Step 1: Finding the number of possible combinations:
To find the number of possible combinations, we can use the concept of permutations. Since we have 4 digits (2, 4, 6, and 8), there are 4 choices for the first digit, 3 choices for the second digit, 2 choices for the third digit, and only 1 choice for the fourth digit. Therefore, the total number of possible combinations is given by:
4 × 3 × 2 × 1 = 24

Step 2: Finding the sum of each digit position:
Now, we need to calculate the sum of each digit position. Since the digits 2, 4, 6, and 8 can appear in any position, we can calculate the sum for each position separately.

Thousands place:
The thousands place can have any of the four digits (2, 4, 6, 8). Therefore, the sum for the thousands place is given by:
(2 + 4 + 6 + 8) × 3! = 20 × 6 = 120

Hundreds place:
Similarly, the hundreds place can have any of the remaining three digits. Therefore, the sum for the hundreds place is given by:
(2 + 4 + 6 + 8) × 2! = 20 × 2 = 40

Tens place:
The tens place can have any of the remaining two digits. Therefore, the sum for the tens place is given by:
(2 + 4 + 6 + 8) × 1! = 20 × 1 = 20

Units place:
Finally, the units place has only one remaining digit. Therefore, the sum for the units place is simply the remaining digit, which is 2.

Step 3: Calculating the total sum:
To calculate the total sum, we need to add up the sums of each digit position:
Total sum = Thousands place sum + Hundreds place sum + Tens place sum + Units place sum
Total sum = 120 + 40 + 20 + 2 = 182

Therefore, the sum of all 4-digit numbers containing the digits 2, 4, 6, and 8 without repetition is 182.

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