Final answer:
The number of integral solutions to the equation a + b + c = 18, where a ≥1, b ≥2, and c ≥3, is found by converting the original equation into a' + b' + c' = 12 and then calculating the number of non-negative integral solutions, which is 14C2, as option d suggests.
Step-by-step explanation:
The question is asking for the number of integral solutions to the equation a + b + c = 18 under the given conditions, which are a ≥ 1, b ≥ 2, and c ≥ 3. To solve this, we should account for the minimum values of a, b, and c by subtracting these from 18, reducing the equation to a' + b' + c' = 12, where a' = a - 1, b' = b - 2, and c' = c - 3 and all are now ≥ 0. The number of non-negative integral solutions is equivalent to the solution of the equation using combinations, specifically 14 choose 2, as there will be 14 items to place into 2 partitions to create 3 groups resembling a', b', and c'.
Using the formula for combinations, the number of solutions is calculated as: 14C2, which is the number of ways to choose 2 elements from a 14-element set without concerning order. Therefore, the answer is option d. 14C2.