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Math127121 assignment 6solutions

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Discrete Mathematics  (Mathematics 1271)

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Academic year: 2021/2022
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Math 1271 Assignment 6 Solutions

  1. Licence plates for a certain province have four letters followed by three digits. How many such plates are possible?

Solution:This is selection with replacement, so we have 26 4 = 456976 ways to choose the letters, and 10 3 = 1000 ways to choose the digits. Thus there are 456976 ×1000 = 456976000 possible plates.

  1. Three dice are rolled. How many outcomes are in the sample space of this experiment? What is the event that the sum of the numbers rolled isten?

Solution:This is a selection with replacement, so we have 6 3 = 216 elements in the sample space. Describe an outcome as an ordered triple (x, y, z),wherexis the num- ber on the first die,ythe number on the second, andzthe number on the third. then the event that the sum rolled is ten is:{(6, 3 ,1),(6, 1 ,3),(6, 2 ,2),(5, 4 ,1),(5, 1 ,4), (5, 3 ,2),(5, 2 ,3),(4, 5 ,1),(4, 1 ,5),(4, 4 ,2),(4, 2 ,4),(4, 3 ,3),(3, 6 ,1),(3, 1 ,6),(3, 5 ,2), (3, 2 ,5),(3, 4 ,3),(3, 3 ,4),(2, 6 ,2),(2, 2 ,6),(2, 5 ,3),(2, 3 ,5),(2, 4 ,4),(1, 6 ,3),(1, 3 ,6), (1, 5 ,4),(1, 4 ,5)}.

  1. How many even positive integers between 100 and 999 inclusive are not divisible by either 5 or 3?

Solution: LetS ={ 100 , 101 , 102 ,· · ·, 999 }, A ={x∈S|x is divisible by 2}, B={x∈S|x is divisible by 5},andC={x∈S|x is divisible by 3}.We are looking for|A(B∪C)|.We haveA(B∪C) =A((A∩B)∪(A∩C)),so|A(B∪ C)|=|A| − |((A∩B)∪(A∩C))|,and|((A∩B)∪(A∩C))|=|A∩B|+|A∩ C| − |A∩B∩C|(by the inclusion/exclusion principle). From the definitions, A∩B={x∈S|x is divisible by 10}, A∩C={x∈S|x is divisible by 6},and A∩B∩C={x∈S|x is divisible by 30}.The smallest multiple of 30 inSis 120. We have 999−120 = 879,and 879 = 30×29 + 9,so|A∩B∩C|= 1 + 29 = 30. Doing similar calculations for the other sets, we get|A∩B|= 90,|A∩C|= 150, and|A|= 450,so|A(B∪C)|=|A| − |((A∩B)∪(A∩C))|=|A| −(|A∩B|+|A∩ C| − |A∩B∩C|) = 450−(90 + 150−30) = 240.

  1. Of 100 personal computer users surveyed, 32 use brand A, 31 use brand B, and 30 use brand C. Twelve use both A and B, fifteen use A and C, and sixteen use B and C. Six use all three. How many use exactly one of A, B, or C? How many only use brands other than A, B, or C?

Solution:LetUdenote the set of all our 100 users, and letAdenote the set of those who use brandA,and similarly forBandC have|A|= 32,|B|= 31, |C|= 30,|A ∩ B|= 12,|A ∩ C|= 15,|B ∩ C|= 16,and|A ∩ B ∩ C|= 6 this we get|A ∪ B ∪ C|=|A|+|B|+|C| − |A ∩ B| − |A ∩ C| − |B ∩ C|+|A ∩ B ∩ C|= 32 + 31 + 30− 12 − 15 −16 + 6 = 56 set of users who use exactly one of A,B,Cis (A ∪ B ∪ C)((A ∩ B)∪(A ∩ C)∪(A ∩ C)).Using inclusion/exclusion, |(A ∩ B)∪(A ∩ C)∪(A ∩ C)|=|A ∩ B|+|A ∩ C|+|B ∩ C| − |A ∩ B ∩ C| − |A ∩ B ∩ C| − |A ∩ B ∩ C|+|A ∩ B ∩ C|= 12 + 15 + 16− 6 − 6 −6 + 6 = 31,so the number of users who use exactly one ofA,B, orC,is 56−31 = 25 set of 1

2

those who use only brands other thatA, B,andCisU(A ∪ B ∪ C).We have |U(A ∪ B ∪ C)|=|U| − |C ∪ B ∪ C|= 100−56 = 44.

  1. SupposeAis a countably infinite set. Show that there exists an injective function f:A→Asuch thatfis not surjective. Show that there exists a surjective function g:A→Athat is not injective.

Solution:SinceAis countably infinite, there exists a bijective functionh:A→N. Define a functionh 1 :N→ Nbyh 1 (n) =n+ 1 1 is injective, since h 1 (n) =h 1 (m) impliesn=m,but it is not surjective, since 0 is not in the range of h 1 .The functionh− 1 ◦h 1 ◦h:A→Ais then injective but not surjective. Define h 2 :N→Nbyh 2 (0) = 0 andh 2 (n) =n−1 forn≥ 1 .Thenh 2 is surjective, since n=h 2 (n+ 1) for anyn,but it is not injective, sinceh 2 (0) =h 2 (1).The function h− 1 ◦h 2 ◦h:A→Ais then surjective but not injective. (Notice that the bijection hallows us to effectively identifyAwithNfor questions of cardinality.)

  1. SupposeAis a countable set, and thatB⊆A thatA\Bis countable.

Solution: It was shown in practice problem set 6, solution to question 6, that subsets of countable sets are countable. SinceAis countable, and (A\B)⊆A,it follows thatA\Bis countable.

  1. SupposeAis a finite set and thatBis a countable set. Show thatA∪Bis countable. Does it follow thatA∪Bis finite?

Solution:SinceAis a finite set, we can arrange it into a finite sequence: A= {a 0 , a 1 ,... , ak}.SinceBis a countable set, we can arrange it into a sequence, possibly finite, possibly infinite:B={b 0 , b 1 , b 2 ,.. .}.We define a sequencecnby cn=anif 0≤n≤k,andcn=bn−k− 1 ifn≥k+ 1 we haveA∪B= {a 0 ,... , ak, b 0 , b 1 , b 2 ,.. .}={c 0 , c 1 , c 2 ,.. .},soA∪Bis countable. IfA=∅and B=N,thenAis finite,Bis countable, butA∪B=B=Nis infinite, so the second statement does not follow.

  1. LetAandBbe two sets. Suppose thatA∪Bis countable. Show thatAandB must both be countable. IfA∪Bwere uncountable, would it follow thatAandB were both uncountable?

Solution:Suppose thatA∪Bis countable. It was shown in practice problem set 6, solution to question 6, that subsets of countable sets are countable. SinceAand Bare both subsets ofA∪B,it follows that they are both countable. IfA=P(N) andB=∅,thenA∪B=A=P(N) which is uncountable, butBis finite, and so countable. Thus the second statement does not follow.

  1. How many ways can you arrange the letters of the word HIPPOPOTAMUS?

Solution: There are twelve letters altogether, but three Ps and two Os. Thus there are 12!/(3!2!) distinct arrangements.

  1. A department has four full professors, two associate professors, and three assistant professors. How many ways can a committee of four be chosen if there must be at least one professor of each rank on the committee?
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Math127121 assignment 6solutions

Course: Discrete Mathematics  (Mathematics 1271)

23 Documents
Students shared 23 documents in this course

University: Lakehead University

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Math 1271 Assignment 6
Solutions
1. Licence plates for a certain province have four letters followed by three digits.
How many such plates are possible?
Solution: This is selection with replacement, so we have 264= 456976 ways to
choose the letters, and 103= 1000 ways to choose the digits. Thus there are
456976 ×1000 = 456976000 possible plates.
2. Three dice are rolled. How many outcomes are in the sample space of this
experiment? What is the event that the sum of the numbers rolled is ten?
Solution: This is a selection with replacement, so we have 63= 216 elements in the
sample space. Describe an outcome as an ordered triple (x, y, z),where xis the num-
ber on the first die, ythe number on the second, and zthe number on the third. then
the event that the sum rolled is ten is: {(6,3,1),(6,1,3),(6,2,2),(5,4,1),(5,1,4),
(5,3,2),(5,2,3),(4,5,1),(4,1,5),(4,4,2),(4,2,4),(4,3,3),(3,6,1),(3,1,6),(3,5,2),
(3,2,5),(3,4,3),(3,3,4),(2,6,2),(2,2,6),(2,5,3),(2,3,5),(2,4,4),(1,6,3),(1,3,6),
(1,5,4),(1,4,5)}.
3. How many even positive integers between 100 and 999 inclusive are not divisible
by either 5 or 3?
Solution: Let S={100,101,102,· · · ,999}, A ={x∈S|x is divisible by 2},
B={x∈S|x is divisible by 5},and C={x∈S|x is divisible by 3}.We are
looking for |A\(B∪C)|.We have A\(B∪C) = A\((A∩B)∪(A∩C)),so |A\(B∪
C)|=|A| − |((A∩B)∪(A∩C))|,and |((A∩B)∪(A∩C))|=|A∩B|+|A∩
C| − |A∩B∩C|(by the inclusion/exclusion principle). From the definitions,
A∩B={x∈S|x is divisible by 10}, A ∩C={x∈S|x is divisible by 6},and
A∩B∩C={x∈S|x is divisible by 30}.The smallest multiple of 30 in Sis 120.
We have 999 −120 = 879,and 879 = 30 ×29 + 9,so |A∩B∩C|= 1 + 29 = 30.
Doing similar calculations for the other sets, we get |A∩B|= 90,|A∩C|= 150,
and |A|= 450,so |A\(B∪C)|=|A| − |((A∩B)∪(A∩C))|=|A| − (|A∩B|+|A∩
C| − |A∩B∩C|) = 450 −(90 + 150 −30) = 240.
4. Of 100 personal computer users surveyed, 32 use brand A, 31 use brand B, and
30 use brand C. Twelve use both A and B, fifteen use A and C, and sixteen use B
and C. Six use all three. How many use exactly one of A, B, or C? How many only
use brands other than A, B, or C?
Solution: Let Udenote the set of all our 100 users, and let Adenote the set of
those who use brand A, and similarly for Band C. We have |A| = 32,|B| = 31,
|C| = 30,|A ∩ B| = 12,|A ∩ C| = 15,|B ∩ C| = 16,and |A ∩ B ∩ C| = 6.From this
we get |A ∪ B ∪ C| =|A| +|B| +|C| − |A ∩ B| − |A ∩ C| − |B ∩ C| +|A ∩ B ∩ C| =
32 + 31 + 30 −12 −15 −16 + 6 = 56.The set of users who use exactly one of
A,B,Cis (A ∪ B ∪ C)\((A ∩ B)∪(A ∩ C)∪(A ∩ C)).Using inclusion/exclusion,
|(A ∩ B)∪(A ∩ C)∪(A ∩ C)|=|A ∩ B| +|A ∩ C| +|B ∩ C| − |A ∩ B ∩ C| − |A ∩
B ∩ C| − |A ∩ B ∩ C| +|A ∩ B ∩ C| = 12 + 15 + 16 −6−6−6 + 6 = 31,so the
number of users who use exactly one of A,B, or C, is 56 −31 = 25.The set of
1

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