MAT2612 Assignment 3 solutions 2026
All questions solved with full working. If need exam help any module reach me
0793226427
ASSIGNMENT 03
Total Marks: 100
Unique no.:202586
ONLY FOR YEAR MODULE
All questions will be marked.
,Question 1
Question 1.1
Question
Twenty cards numbered 1 to 20 are placed face down on a table. Cards are selected
one at a time and turned over. If two of the selected cards add up to 21, the player
loses.
Use the pigeonhole principle to show that if 11 cards are chosen, then the player
can never win the game. State clearly what the pigeons and the pigeonholes are.
Form the pairs
The numbers from 1 to 20 can be grouped into pairs whose sum is 21.
(1, 20)
(2, 19)
(3, 18)
(4, 17)
(5, 16)
(6, 15)
(7, 14)
(8, 13)
(9, 12)
(10, 11)
There are therefore 10 pairs.
These pairs are the pigeonholes.
Identify the pigeons
The 11 selected cards are the pigeons.
So we have
• Number of pigeons = 11
• Number of pigeonholes = 10
, Apply the Pigeonhole Principle
The pigeonhole principle states:
If more than 𝑛objects are placed into 𝑛boxes, then at least one box must contain two
objects.
Here,
• 11 cards are placed into 10 pairs.
Since
11 > 10,
at least one pair must contain both cards.
Every pair sums to 21.
Therefore, among any 11 selected cards there must be two cards whose sum is 21.
Hence the player loses.
Conclusion
Since every selection of 11 cards must contain one complete pair whose numbers add
up to 21, it is impossible for the player to win after choosing 11 cards.
Pigeons
The 11 selected cards.
Pigeonholes
The 10 pairs
{1,20}, {2,19}, {3,18}, {4,17}, {5,16}, {6,15}, {7,14}, {8,13}, {9,12}, {10,11}.
Question 1.2
Question
A store has an introductory sale on 12 types of candy bars.
A customer chooses one bar from any five different types, and the total cost is at
most R1.75.
Use the extended pigeonhole principle to show that although different selections may
cost different amounts, there must be at least two different selections that have the
same total cost.
All questions solved with full working. If need exam help any module reach me
0793226427
ASSIGNMENT 03
Total Marks: 100
Unique no.:202586
ONLY FOR YEAR MODULE
All questions will be marked.
,Question 1
Question 1.1
Question
Twenty cards numbered 1 to 20 are placed face down on a table. Cards are selected
one at a time and turned over. If two of the selected cards add up to 21, the player
loses.
Use the pigeonhole principle to show that if 11 cards are chosen, then the player
can never win the game. State clearly what the pigeons and the pigeonholes are.
Form the pairs
The numbers from 1 to 20 can be grouped into pairs whose sum is 21.
(1, 20)
(2, 19)
(3, 18)
(4, 17)
(5, 16)
(6, 15)
(7, 14)
(8, 13)
(9, 12)
(10, 11)
There are therefore 10 pairs.
These pairs are the pigeonholes.
Identify the pigeons
The 11 selected cards are the pigeons.
So we have
• Number of pigeons = 11
• Number of pigeonholes = 10
, Apply the Pigeonhole Principle
The pigeonhole principle states:
If more than 𝑛objects are placed into 𝑛boxes, then at least one box must contain two
objects.
Here,
• 11 cards are placed into 10 pairs.
Since
11 > 10,
at least one pair must contain both cards.
Every pair sums to 21.
Therefore, among any 11 selected cards there must be two cards whose sum is 21.
Hence the player loses.
Conclusion
Since every selection of 11 cards must contain one complete pair whose numbers add
up to 21, it is impossible for the player to win after choosing 11 cards.
Pigeons
The 11 selected cards.
Pigeonholes
The 10 pairs
{1,20}, {2,19}, {3,18}, {4,17}, {5,16}, {6,15}, {7,14}, {8,13}, {9,12}, {10,11}.
Question 1.2
Question
A store has an introductory sale on 12 types of candy bars.
A customer chooses one bar from any five different types, and the total cost is at
most R1.75.
Use the extended pigeonhole principle to show that although different selections may
cost different amounts, there must be at least two different selections that have the
same total cost.
