A bus driver pays all tolls, using only nickels and dimes, by throwing one coin at a time into the mechanical toll collector.

a) Find a recurrence relation for the number of different ways the bus driver can pay a toll of n cents (where the order in which the coins are used matters).

b) In how many different ways can the driver pay a toll of 45 cents.

Answer: a) An = An-1 + An-2

b) 55ways

Step-by-step explanation:

a) a nickel is 5 cents and a dime is 10cent so a multiple of 5 cents is the possible way to pay the tolls in both choices.

Let An represents the number of possible ways the driver can pay a toll of 5n cents, so that

An = 5n cents

Case 1: Using a nickel for payment which is 5 cents, the number of ways given as;

An-1 = 5 (n-1)

Case 2: using a dime which is two 5 cents, the number of ways is given as;

An-2 = 5 (n-2)

Summing up the number of ways, we have

An = An-1 + An-2

From the relation,

If n = 0, Ao = 1

n = 1, A1 = 1

b) 45 cents paid in multiples of 5cents will give us 9 ways (A9)

From the relation, we have that

Ao = 1

A1 = 1

An = An-1 + An-2

Ao = 1

A1 = 1

A2 = A1+Ao = 1+1 = 2

A3 = A2 + A1 = 3

A4 = A3+A2=5

A5=A4+A3=8

A6=A5+A4=13

A7 = A6+A5 = 21

A8 = A7+A6 = 34

A9 = A8+A7 = 55

So there are 55ways to pay 45cents.