A owes B Rs 50. He agrees to pay B over a number of consecutive days starting…

When a fixed amount must be paid using notes of only two denominations, one note per day, the number of distinct day-by-day repayment sequences equals the number of distinct arrangements of a multiset: for a set of n days where one denomination is used p times and the other q times (p + q = n), the count of distinct orderings is n!/(p! . q!). Every valid split of the two denominations that adds up to the target amount is a separate case, and the sequence-counts of all valid cases must be added together.

1. Let a = number of Rs 10 notes used and b = number of Rs 20 notes used. Since the total is Rs 50: 10a + 20b = 50, i.e. a + 2b = 5, with a, b >= 0 (integers).

2. The non-negative integer solutions of a + 2b = 5 are: b = 0 (a = 5), b = 1 (a = 3), and b = 2 (a = 1). (b = 3 would need a = -1, which is invalid.)

3. For each solution, the number of days is a + b, and the number of distinct day-by-day sequences is (a + b)! / (a! . b!): for a = 5, b = 0 -> 5 days -> 1 sequence; for a = 3, b = 1 -> 4 days -> 4 sequences; for a = 1, b = 2 -> 3 days -> 3 sequences.

4. Adding the sequence-counts of all three valid cases: 1 + 4 + 3 = 8 distinct ways.

An equivalent check: in each case, choose which of the a + b days carry the Rs 20 note - this gives C(5,0), C(4,1), and C(3,2) respectively for the three cases above. These combination counts also sum to 1 + 4 + 3 = 8, confirming the total of 8 distinct repayment sequences.