UNIVERSITY OF SOUTH AFRICA
College of Science, Engineering and Technology
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
MAT2612: Discrete Mathematics
Assignment 03 | Year Module, 2026
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
MAT2612
Module Code:
Discrete Mathematics
Module Name:
Pigeonhole Principle and Relations
Assignment Topic:
03
Assignment Number:
202586
Unique Number:
100
Total Marks:
Submitted in partial fulfilment of the requirements for MAT2612, UNISA 2026
, UNISA | MAT2612 Assignment 03
Question 1
1.1
Twenty cards numbered 1 to 20 are placed face down on a table. Cards are se-
lected one at a time and turned over, and the player loses if two of the selected
cards add up to 21. Us the pigeonhole principle to show that if 11 cards are cho-
sen, then the player can never win the game. State clearly what the pigeons and
the pigeonholes are.
The twenty numbers can be grouped into pairs that each add up to 21:
(1, 20), (2, 19), (3, 18), (4, 17), (5, 16), (6, 15), (7, 14), (8, 13), (9, 12), (10, 11).
This gives exactly 10 pairs, and together they account for all twenty numbers from 1 to 20.
The pigeons are the 11 selected cards. The pigeonholes are the 10 pairs
{1, 20}, {2, 19}, {3, 18}, . . . , {10, 11}.
Since 11 cards are chosen and there are only 10 pigeonholes, the pigeonhole principle guaran-
tees that at least two of the selected cards fall into the same pigeonhole, that is, at least two
selected cards form one of the pairs above. For every such pair the sum is 21:
1 + 20 = 2 + 19 = 3 + 18 = · · · = 10 + 11 = 21.
Therefore at least two of the 11 selected cards must add up to 21.
If 11 cards are chosen, the player can never win.
Page 2 of 10
College of Science, Engineering and Technology
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
MAT2612: Discrete Mathematics
Assignment 03 | Year Module, 2026
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
MAT2612
Module Code:
Discrete Mathematics
Module Name:
Pigeonhole Principle and Relations
Assignment Topic:
03
Assignment Number:
202586
Unique Number:
100
Total Marks:
Submitted in partial fulfilment of the requirements for MAT2612, UNISA 2026
, UNISA | MAT2612 Assignment 03
Question 1
1.1
Twenty cards numbered 1 to 20 are placed face down on a table. Cards are se-
lected one at a time and turned over, and the player loses if two of the selected
cards add up to 21. Us the pigeonhole principle to show that if 11 cards are cho-
sen, then the player can never win the game. State clearly what the pigeons and
the pigeonholes are.
The twenty numbers can be grouped into pairs that each add up to 21:
(1, 20), (2, 19), (3, 18), (4, 17), (5, 16), (6, 15), (7, 14), (8, 13), (9, 12), (10, 11).
This gives exactly 10 pairs, and together they account for all twenty numbers from 1 to 20.
The pigeons are the 11 selected cards. The pigeonholes are the 10 pairs
{1, 20}, {2, 19}, {3, 18}, . . . , {10, 11}.
Since 11 cards are chosen and there are only 10 pigeonholes, the pigeonhole principle guaran-
tees that at least two of the selected cards fall into the same pigeonhole, that is, at least two
selected cards form one of the pairs above. For every such pair the sum is 21:
1 + 20 = 2 + 19 = 3 + 18 = · · · = 10 + 11 = 21.
Therefore at least two of the 11 selected cards must add up to 21.
If 11 cards are chosen, the player can never win.
Page 2 of 10
