MAT 226 Discrete Mathematics Homework Assignment 7 Solutions

MAT 226 Discrete Mathematics
Homework Assignment 7 Solutions
1. Find a recurrence relation for the number of permutations of a set with n elements.
(Don’t forget to include a statement about the valid values of n, as well as the initial
conditions.)
Let an represent the number of permutations of a set with n elements. Then the
recurrence relation is give by an = n*an-1 for n ≥ 1 and a0 = 1.
2. Find a recurrence relation for the number of bit strings of length n that contain an
even number of 0s.
Let an be the number of bit strings of length n that contain an even number of 0s.
Then there are two mutually exclusive cases for an: If the first digit is a 1, then an =
an-1. If the first digit is a 0, then an is given by the number of bit strings of length n – 1
that contain an odd number of 0s, which is 2n-1 - an-1. So, an = an-1 + (2n-1 - an-1) =
2n-1 for n ≥ 2.
3. A bus driver pays all tolls using only nickels and dimes, by throwing one coin at a
time into the mechanical toll collector. (All tolls are a multiple of 5 cents.)
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.
Let an be the number of ways to pay a toll of 5n cents. Then, the bus driver can
either put in a nickel and pay the remainder of the toll in an-1 ways or put in a
dime and pay the remainder of the toll in an-2 ways. So, an = an-1 + an-2 for n ≥ 2
and a0 = a1 = 1.
b) In how many different ways can the driver pay a toll of 45 cents?
a0 = a1 = 1, a1 = 2, a1 = 3, a1 = 5, a1 = 8, a1 = 13, a1 = 21, a1 = 34, and a1 = 55.
4. A nuclear reactor has created 18 grams of a particular radioactive isotope. Every
hour 1% of this radioactive isotope decays. Set up a recurrence relation for the
amount of this isotope left n hours after its creation. How much of the isotope is left
after five hours?
Let an be the amount of the isotope left n hours after its creation. Then an = 0.99*an-1
for n ≥ 1 and a0 = 18. Therefore (to the nearest hundredth), a1 = 17.82, a2 = 17.64,
a3 = 17.47, a4 = 17.29, and a5 = 15.56.