TEST 1 PRACTICE PROBLEMS EXAM
2025 QUESTIONS AND ANSWERS
In a competition between players X and Y, the first player to win three games in a row or a total
of four games wins. How many ways can the competition be played if X wins the first game and
Y wins the second and third games? (Draw a tree.) - ANS 7 ways.
See D2L for the tree.
We usually write numbers in decimal form (or base 10), meaning numbers are composed using
10 different "digits" 0, 1, ... , 9. Sometimes though it is useful to write numbers hexadecimal or
base 16. Now there are 16 distinct digits that can be used to form numbers: 0, 1, ... , 9, A, B, C,
D, E, F, G. So for example, a 3 digit hexadecimal number might be 2B8.
(a) How many 2-digit hexadecimals are there in which the first digit is E or F? Explain your
answer in terms of the additive principle (using either events or sets)
(b) Explain why your answer to the previous part is correct in terms of the multiplicative
principle (using either events or sets). Why do both the additive and multiplicative principles
give you the same answer?
(c) How many 3-digit hexadecimals start with a letter (A-F) and end with a numeral (0-9)?
Explain.
1
(d) How many 3-digit hexadecimals start with a letter (A-F) or end with a numeral (0-9) -
Page
ANS *(a)*
@COPYRIGHT SIRJOEL 2025/2026
, Consider two cases:
i) first letter is fixed to E: there are two positions to fill: The first one can be filled in one way: E.
The second one can be filled in 16 different ways (0-9 and A-F). So we have 1x16 different ways.
ii) first letter is fixed to F: there are two positions to fill: The first one can be filled in one way: F.
The second one can be filled in 16 different ways (0-9 and A-F). So we have 1x16 different
ways.
By addition rule, we need to add these two: 1x16 + 1x16 = 2
*(b)*
There are two digits i.e. there are two positions to fill: The first one can be filled in two different
ways: E or F. The second one can be filled in 16 different ways (0-9 and A-F). So we have 2x16
different ways
*(c)*
There are three digits i.e. there are three positions to fill: The first one can be filled in 6
different ways: A-F. The second one can be filled in 16 different ways (0-9 and A-F). The third
one can be filled in 10 different ways (0-9). So we have 6x16x10 different ways
*(d)*
We must use inclusion-exclusion formula: Let M be the set of 3-digit hexadecimals start with a
letter (A-F). Let N be the set of 3-digit hexadecimals that end with a numeral (0-9). First we
calculate number of elements of M and N:
First M: There are three digits i.e. there are three positions to fill: The first one can be filled in 6
different ways: A-F. The second and third ones can be filled in 16 different ways (0-9 and AF). So
we have 6x16x16 different ways.
2
Page
@COPYRIGHT SIRJOEL 2025/2026
2025 QUESTIONS AND ANSWERS
In a competition between players X and Y, the first player to win three games in a row or a total
of four games wins. How many ways can the competition be played if X wins the first game and
Y wins the second and third games? (Draw a tree.) - ANS 7 ways.
See D2L for the tree.
We usually write numbers in decimal form (or base 10), meaning numbers are composed using
10 different "digits" 0, 1, ... , 9. Sometimes though it is useful to write numbers hexadecimal or
base 16. Now there are 16 distinct digits that can be used to form numbers: 0, 1, ... , 9, A, B, C,
D, E, F, G. So for example, a 3 digit hexadecimal number might be 2B8.
(a) How many 2-digit hexadecimals are there in which the first digit is E or F? Explain your
answer in terms of the additive principle (using either events or sets)
(b) Explain why your answer to the previous part is correct in terms of the multiplicative
principle (using either events or sets). Why do both the additive and multiplicative principles
give you the same answer?
(c) How many 3-digit hexadecimals start with a letter (A-F) and end with a numeral (0-9)?
Explain.
1
(d) How many 3-digit hexadecimals start with a letter (A-F) or end with a numeral (0-9) -
Page
ANS *(a)*
@COPYRIGHT SIRJOEL 2025/2026
, Consider two cases:
i) first letter is fixed to E: there are two positions to fill: The first one can be filled in one way: E.
The second one can be filled in 16 different ways (0-9 and A-F). So we have 1x16 different ways.
ii) first letter is fixed to F: there are two positions to fill: The first one can be filled in one way: F.
The second one can be filled in 16 different ways (0-9 and A-F). So we have 1x16 different
ways.
By addition rule, we need to add these two: 1x16 + 1x16 = 2
*(b)*
There are two digits i.e. there are two positions to fill: The first one can be filled in two different
ways: E or F. The second one can be filled in 16 different ways (0-9 and A-F). So we have 2x16
different ways
*(c)*
There are three digits i.e. there are three positions to fill: The first one can be filled in 6
different ways: A-F. The second one can be filled in 16 different ways (0-9 and A-F). The third
one can be filled in 10 different ways (0-9). So we have 6x16x10 different ways
*(d)*
We must use inclusion-exclusion formula: Let M be the set of 3-digit hexadecimals start with a
letter (A-F). Let N be the set of 3-digit hexadecimals that end with a numeral (0-9). First we
calculate number of elements of M and N:
First M: There are three digits i.e. there are three positions to fill: The first one can be filled in 6
different ways: A-F. The second and third ones can be filled in 16 different ways (0-9 and AF). So
we have 6x16x16 different ways.
2
Page
@COPYRIGHT SIRJOEL 2025/2026
