Final answer:
By analyzing the given sum of the numbers formed with the digits 1, 2, x, and y, and recognizing that each digit contributes equally across all combinations, we can calculate that x + y equals 13. Therefore, the answer is option C: 11.
Step-by-step explanation:
The question concerns a combinatorics and algebra problem in which we are given 4 distinct digits, 1, 2, x, and y, to form natural numbers with 3 different digits. We are told there are 18 such numbers summing up to 4508, and we need to calculate x + y. To solve this problem, we first recognize that each digit appears an equal number of times in each position when all different possible combinations are made.
Since there are 4 unique digits and we want to form 3-digit numbers, each digit will appear in the hundreds place (4-1)! times, in the tens place (4-1)! times, and in the ones place (4-1)! times. This is a total of 6 occurrences per position, which gives 6+6+6=18 occurrences in total for each digit across all 18 numbers. Hence, every digit contributes an equal part to the total sum of the numbers.
By taking 1 and 2, we can calculate their total contribution as 1+2=3. Since this contribution is found in three places of each of the 18 numbers, we multiply by 18, and then by 3 (places), resulting in a subtotal of 162. Now, we subtract this subtotal from the given total sum: 4508 - 162 = 4346. The remaining sum must be the total contribution from x and y. Dividing this by 18 and then by 3 gives us the sum of x + y: 4346 / 54 = 80 / 6 = 13.33. Since we are dealing with natural digits, we round down to the nearest whole number, which is 13. Thus, x + y = 13, which corresponds to option C: 11.