Correct Option (C)

The problem requires determining the number of distinct combinations of ₹1, ₹10, and ₹50 denominations that sum up to ₹107. Let x be the number of ₹50 notes, y be the number of ₹10 notes, and z be the number of ₹1 notes. The objective is to find the number of non-negative integer solutions to the equation: 50x + 10y + z = 107.

We can systematically enumerate the possibilities by considering the maximum possible number of ₹50 notes, then ₹10 notes, and finally determining the required ₹1 notes.

Case 1: Using 0 notes of ₹50 (x = 0)
The equation becomes 10y + z = 107. Since z ≥ 0, 10y ≤ 107. Thus, y can take integer values from 0 to 10. For each value of y, z is uniquely determined (z = 107 - 10y). This yields 11 distinct ways (from y=0, z=107 to y=10, z=7).
Case 2: Using 1 note of ₹50 (x = 1)
The equation becomes 50(1) + 10y + z = 107, which simplifies to 10y + z = 57. Since z ≥ 0, 10y ≤ 57. Thus, y can take integer values from 0 to 5. For each value of y, z is uniquely determined (z = 57 - 10y). This yields 6 distinct ways (from y=0, z=57 to y=5, z=7).
Case 3: Using 2 notes of ₹50 (x = 2)
The equation becomes 50(2) + 10y + z = 107, which simplifies to 100 + 10y + z = 107, or 10y + z = 7. Since z ≥ 0, 10y ≤ 7. Thus, y can only be 0. For y=0, z = 7. This yields 1 distinct way (2 notes of ₹50, 0 notes of ₹10, 7 notes of ₹1).
Case 4: Using 3 or more notes of ₹50 (x ≥ 3)
If x = 3, the value would be 50(3) = 150, which exceeds ₹107. Therefore, no further combinations are possible.

Summing the ways from each case: 11 (Case 1) + 6 (Case 2) + 1 (Case 3) = 18 ways.

Incorrect Options:

Options A (16), B (17), and D (19) are incorrect because a comprehensive enumeration of all possible combinations of ₹1, ₹10, and ₹50 denominations that sum to ₹107, as detailed above, yields a total of 18 distinct ways. These options do not reflect the complete set of valid payment methods.