MATH2111
RMIT
TUTORIAL 9
1. A person buying a personal computer system is offered a choice of three
models of the basic unit, two models of keyboard, and two models of printer.
How many distinct systems can be purchased?
3
𝐶1 × 2𝐶1 × 2𝐶1 = 3 × 2 × 2 = 12
2. Suppose that a code consists of five characters, two letters followed by three
digits. Find the number of:
(a) codes;
26
𝑃1 × 26𝑃1 × 10𝑃1 × 10𝑃1 × 10𝑃1
= 26𝐶1 × 26𝐶1 × 10𝐶1 × 10𝐶1 × 10𝐶1
= 26 × 26 × 10 × 10 × 10
= 676,000
(b) codes with distinct letter.
26
𝑃2 × (10)3 =
26
𝑃1 × 25𝑃1 × (10)3 =
26 × 25 × 24!
× 1000 = 650,000
(26 − 2)!
3. Consider all positive integers with three digits. (Note that zero cannot be the
first digit.) Find the number of them which are:
(a) greater than 700;
3
𝑃1 × 10 × 10 − 1 = 299
(b) odd;
9
𝑃1 × 10𝑃1 × 5𝑃1 = 450
(c) divisible by 5.
9
𝑃1 × 10𝑃1 × 2𝑃1 = 180
4. The letters ABCDE are to be used to form strings of length 3. How many
strings:
(a) begin with A, allowing repetitions
1
𝑃1 × 5𝑃1 × 5𝑃1 = 25
(b) begin with A, repetitions are not allowed?
1
𝑃1 × 4𝑃2 = 1𝑃1 × 4𝑃1 × 3𝑃1 = 1𝑃1 × 4𝐶1 × 3𝐶1 = 12
5. Suppose that each child born is equally likely to be a boy or a girl. Consider a
family with exactly three children.
(a) List the elements in the sample space.
𝑆 = {𝑏𝑏𝑏, 𝑏𝑏𝑔, 𝑏𝑔𝑏, 𝑔𝑏𝑏, 𝑏𝑔𝑔, 𝑔𝑏𝑔, 𝑔𝑔𝑏, 𝑔𝑔𝑔}
(b) What is the probability that the family has exactly one girl?
MATH2111
RMIT
3
8
(c) What is the probability that at least two children are girls?
3 1 1
𝑃(𝑔 = 3) + 𝑃(𝑔 = 3) = + =
8 8 2
𝑃(𝑔 = 1) =
6. Hexadecimal numbers are made using the sixteen digits 0, 1, 2, 3, 4, 5, 6, 7, 8,
9, A, B, C, D, E, F. They are denoted by the subscript 16.
(a) How many hexadecimal numbers begin with one of the digits 3 through B, end
with one of the digits 5 through F and are 5 digits long?
9
𝑃1 × (16)3 × 11𝑃1 = 9 × (16)3 × 11
(b) How many strings of hexadecimal digits consist of from one through three
digits?
1 digit → 16
2 digit→ 16 × 16
3 digit→ 16 × 16 × 16
Total strings available = 16 + (16 × 16) + (16 × 16 × 16) =
7. Prove that for all integers n ≥ 3, 𝑛+1𝑃3 − 𝑛𝑃3 = 3 𝑛𝑃2.
(𝑛 + 1)!
𝑛!
𝑛+1
𝑃3 − 𝑛𝑃3 =
−
(𝑛 + 1 − 3)! (𝑛 − 3)!
(𝑛 + 1)(𝑛)(𝑛 − 1)(𝑛 − 2)! (𝑛)(𝑛 − 1)(𝑛 − 2)(𝑛 − 3)!
=
−
(𝑛 − 2)!
(𝑛 − 3)!
= (𝑛)(𝑛 − 1)(𝑛 + 1 − 𝑛 + 2)
3(𝑛)(𝑛 − 1)(𝑛 − 2)!
=
= 3 𝑛𝑃2
(𝑛 − 2)!
8. A group of eight people are going to watch a movie together. Two of the eight
must sit side-by-side. In how many ways can the eight be seated together in a
row?
7
𝑃7 × 2𝑃2 = 10,080
9. Determine the formula to obtain the number of elements in the union of 4
finite sets A, B, C and D. Note that this formula is a specific case of the
inclusion/exclusion principle.