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CSci/Math2112
(9)
Assignment 2
Due May 29, 2015
1. Negate the following statements:
(a) If for the integers x, y, z we know that x divides y and y divides z, then x divides z.
Solution: There exist integers x, y, and z such that x divides y and y divides z and x does not
divide z.
(b) Today is Wednesday and it will be sunny or cloudy.
Solution: Today is not Wednesday or it will not be sunny and not cloudy.
(c) One out of every 10 seeds sprouts.
Solution: The solution to this depends on how you interpret the sentence:
(1) At least one out of every 10 seeds sprouts.
Let S be the set of seeds and A = {B ⊆ S | |B| = 10}. The sentence in symbolic form is thus:
∀N ∈ A∃x ∈ N : x sprouts.
Thus negated: There exists a set of 10 seeds such that all of these do not sprout.
(2) Exactly one out of every 10 seeds sprouts.
The negations for this one is: There exists a set of 10 seeds such that all of these do not sprout or
at least two sprout.
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2. For each pair of statements below, give at least one example of a statement(s) which if added to the first
allows you to deduce the second.
(a) (i) Every integer x is even or odd.
(ii) 2y is even.
Solution: 2y is not odd.
(b) (i) a ∈ A and a ∈ B.
(ii) a ∈ A.
Solution: Any true statement is correct for this since (i) already implies (ii).
(c) (i) We have class today.
(ii) Today is a Monday or Friday.
Solution: We have class on Mondays and Fridays.
or
We have class on Mondays, Wednesdays, and Fridays, and today is not Wednesday.
(2)
3. (BoP 3.1 #4) Five cards are dealt of the standard 52-card deck and lined up in a row. How many such
line-ups are there in which all 5 cards are of the same suit?
Solution: There are 52 choices for the first card. Then the suit is fixed, and there are 12 cards left
from that suit. Thus there are 12 choices for the second card, 11 for the third,... Thus the total number
of line-ups is
52 · 12 · 11 · 10 · 9 = 617760.
Alternative argument idea: There are 4 choices for the suit, and 13 cards in each suit.
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4. (BoP 3.1 #10a/c) This problem concerns lists made from the letters A, B, C, D, E, F, G, H, I, J. How
many length-5 lists can be made from these letters if repetition is not allowed, the list must begin with
a vowel, and the list must contain exactly one A?
Solution: There are two cases to consider:
(i) The first letter is the A. Then there are 9 choices for the second letter, 8 for the third,...
(ii) The first letter is not A. Then there are 4 choices to place the letter A, and 2 choices for the first
letter (E or I). Of the remaining empty spots, there are 8 choices for the first, 7 for the second, and 6
for the third.
Thus the total number of different lists is
9 · 8 · 7 · 6 + 4 · 2 · 8 · 7 · 6 = 2688.
(1)
5. (BoP 3.2 #4) Using only pencil and paper, find the value of
Solution:
(4)
100!
.
95!
100!
= 100 · 99 · 98 · 97 · 96 = 9034502400
95!
6. (BoP 3.2 #8) Compute how many 7-digit numbers can be made from the digits 1, 2, 3, 4, 5, 6, 7 if there
is no repetition and the odd digits must appear in an unbroken sequence.
Solution: Since there are 4 odd digits, the sequence of odd digits can begin on the first empty spot,
the second, the third, or fourth, so there are 4 choices of where to have the first odd digit.
There are 4! ways to arrange the odd digits, and 3! to arrange the even digits. So in total there are
4 · 4! · 3! = 576
such arrangements.
(3)
7. (BoP 3.3 #2) Suppose A is a set for which |A| = 100. How many subsets of A have 5 elements? How
many subsets have 10 elements? How many have 99 elements?
=
Solution: 5 elements: 100
5
100
elements: 100
=
=
100
99
1
(6)
100!
95!5!
= 75287520 10 elements:
100
10
=
100!
90!10!
= 17310309456440 99
8. (BoP 3.3 #8) This problem concerns lists made from the symbols A, B, C, D, E, F, G, H, I.
(a) How many length-5 lists can be made if repetition is not allowed and the list is in alphabetical
order?
Solution:
9
= 126
5
(b) How many length-5 lists can be made if repetition is not allowed and the list is not in alphabetical
order?
Solution: Number of total lists in any order:
9!
(9−5)!
Thus number of lists not in alphabetical order:
9!
9
−
= 14994
4!
5
(3)
9. (BoP 3.3 #10) A department consists of 5 men and 7 women. From this department you select a
committee with 3 men and 2 women. In how many ways can you do this?
Solution:
5
7
·
= 210
3
2