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MAT2612 Assignment 3 Complete Solutions Unisa 2024 Discrete Mathematics Due date 22 July 2024 R150,00   Add to cart

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MAT2612 Assignment 3 Complete Solutions Unisa 2024 Discrete Mathematics Due date 22 July 2024

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MAT2612 Assignment 3 Complete Solutions Unisa 2024 Discrete Mathematics Due date 22 July 2024

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  • July 19, 2024
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MAT2612
ASSIGNMENT 3
FULL SOLUTIONS
Unique Number: 550743
DISCRETE MATHEMATICS
Due date: 22 JULY 2024

UNISA
2024

,SOLUTION:

1.1 Let us create the set of two cards which add up to 21.

S = {(1,20),(2,19),(3,18),(4,17),(5,16),(6,15),(7,14),(8,13),(9,12),(10,11)}

(we are not taking next (11,10) because it is same as (10,11))

Clearly number of elements in set S is |S| = 10

(So now let us choose the cards avoiding sum = 21 , after 10 cards we would have
chosen one card from each element of S , Example (1,2,18,4,16,15,7,8,12,10) and now
we have to choose one more but now we have no option , we have to choose from one
of the elements of S from which we have already taken out one number , so we cannot
avoid a sum of 21.)

Formal Proof :

S is the pigenhole and number of pigeonholes = 10 and 11 numbers which we will
choose will be our pigeons

Clearly , using Pigeonhole principle , we can say that two pigeons will belong to same
pigeonhole or sum of two numbers will be equal to 21.

,

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