Suppose you have sufficient amount of rupee currency in three denominations : ₹ 1, ₹ 10 and ₹ 50. In how many different ways can you pay a bill of ₹ 107 ?

• Let the number of ₹50, ₹10, and ₹1 notes be x, y, and z respectively, satisfying the equation 50x + 10y + z = 107.

• Since z must be non-negative, the maximum value for x is 2 (as 3 * 50 = 150 > 107). Thus, x can be 0, 1, or 2.

• For x = 2: 100 + 10y + z = 107 implies 10y + z = 7. Since z < 10, y must be 0, giving z = 7. This yields 1 way.

• For x = 1: 50 + 10y + z = 107 implies 10y + z = 57. Since z < 10, 10y must be 50, so y = 5 and z = 7. This yields 1 way.

• For x = 0: 10y + z = 107. Since z < 10, 10y must be 100, so y = 10 and z = 7. This yields 1 way.

• Wait, the previous logic assumed z is determined by the remainder. Let's re-evaluate. z = 107 - 50x - 10y. For a valid solution, z must be >= 0. Also, since we are counting ways to pay, z is just the number of 1-rupee notes. The constraint is simply that z >= 0. There is no upper limit on z other than the total amount. However, usually in these problems, 'ways' implies distinct combinations of coins. Let's look at the equation 50x + 10y + z = 107. z is uniquely determined by x and y. So we just need to count the number of non-negative integer pairs (x, y) such that 50x + 10y <= 107.

• If x=0, 10y <= 107 => y can be 0, 1, ..., 10. (11 values)

• If x=1, 50 + 10y <= 107 => 10y <= 57 => y can be 0, 1, ..., 5. (6 values)

• If x=2, 100 + 10y <= 107 => 10y <= 7 => y can be 0. (1 value)

• Total ways = 11 + 6 + 1 = 18.