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Nickels are one of the two types of coins that the bus driver can use to pay the toll.
Each nickel is worth 5 cents.
When considering the number of ways to pay a certain amount using nickels, it’s important to understand that the total value must be a multiple of 5.
For example, if the driver needs to pay 15 cents, they can use three nickels (3 * 5 = 15).
The use of nickels can be combined with dimes to form different combinations.
This will be important when forming the recurrence relation, as you'll see in the subsequent sections.

Dimes are the other type of coin available to the bus driver for toll payments. Each dime is worth 10 cents.
Detecting patterns with dimes is a bit different than with nickels because the value increments in steps of 10.
For example, to make 20 cents, the driver could use two dimes, or four nickels, or a combination of both.
By understanding how dimes incrementally increase the total toll payment value, we integrate this knowledge into forming the recurrence relation.
This highlights our next section where we consider all possible ways to combine these increments.

The core challenge is finding how many different ways the bus driver can pay a specific toll using nickels and dimes.
A useful approach is the recurrence relation. The key insight is recognizing that each toll amount can be reached by adding either a nickel or a dime to smaller toll amounts.
We can express this mathematically as: \[ a(n) = a(n - 5) + a(n - 10) \] This recurrence relation expresses that the number of ways to pay \( n \) cents (denoted as \( a(n) \)) can be obtained by summing the number of ways to pay \( n - 5 \) cents (by adding a nickel) and \( n - 10 \) cents (by adding a dime).
Using this relation, combined with base cases like \( a(0) = 1 \), \( a(5) = 1 \), and \( a(10) = 2 \), we can calculate the total number of ways to pay higher amounts step-by-step.
This is essential for solving both small and large problems, such as finding the specific number of ways to pay 45 cents.