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.
,
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.
,
