a, b and c are natural numbers and a ≤ 50, b ≤ 15, c ≤ 15. Find the number of solutions (a, b, c) of the equation a – b – c = 30.

To solve this problem, we need to find the number of solutions (a, b, c) for the equation a - b - c = 30 where a, b, and c are natural numbers and a ≤ 50, b ≤ 15, c ≤ 15.

Step 1: Rewrite the equation We can rewrite the equation as a = 30 + b + c. This means that the sum of b and c plus 30 must be equal to a.

Step 2: Determine the possible values for b and c Since b and c are natural numbers and each is less than or equal to 15, the smallest possible sum of b and c is 0 (when both are 0) and the largest possible sum is 30 (when both are 15).

Step 3: Determine the possible values for a Given that a = 30 + b + c, the smallest possible value for a is 30 (when b and c are both 0) and the largest possible value for a is 60 (when b and c are both 15). However, since a is also given to be less than or equal to 50, the actual largest possible value for a is 50.

Step 4: Count the number of possible solutions We now need to count the number of possible solutions (a, b, c). For each possible value of a (from 30 to 50), there are a - 30 + 1 possible combinations of values for b and c (since b and c can be any natural numbers that sum to a value between 0 and a - 30). Therefore, the total number of solutions is the sum of a - 30 + 1 for a from 30 to 50.

This can be calculated as follows:

Sum = (30 - 30 + 1) + (31 - 30 + 1) + ... + (50 - 30 + 1) = 1 + 2 + ... + 21 = 231

Therefore, there are 231 possible solutions (a, b, c) for the given equation under the given conditions.