Set operations: Sets resource page.
Note: Using {} to denote the empty set, emptySet.
Page 137 #4 a) {} elementOf {} FALSE
b) {} properSubsetOf {} FALSE
c) {} subsetOf {} TRUE
d) {} elementOF {{}} TRUE
e) {} properSubsetOf {{}} TRUE
f) {} subsetOf {{}} TRUE
VIP: Note that subsetOf is analogous to <=
and properSubsetOf is analogous to <
Page 137 #8 For A = {1,2,3,4,5,6,7}, determine the number of:
7
a. subsets of A 2 = 128
7
b. nonempty subsets of A 2 - 1 = 127
c. proper subsets of A 128-1 = 127
d. nonempty proper subsets of A 128-2 = 126
7*6*5
e. subsets of A containing three elements C(7,3) = ----- = 7*5 = 35
3*2*1
f. subsets of A containing 1, 2
5
How many subsets of {3,4,5,6,7} are possible? 2 = 32
Combining the elements 1, 2 to each of these 32 subsets, gives
us exactly 32 distinct subsets that all contain 1 and 2.
Interesting to think about:
There are 32 subsets of {3, 4, 5, 6, 7}
There are 32 subsets of {3, 4, 5, 6, 7} that contain 1 and 2
There are 32 subsets of {3, 4, 5, 6, 7} that contain 1, but not 2
There are 32 subsets of {3, 4, 5, 6, 7} that contain 2, but not 1
----
128 subsets of {1,2,3,4,5,6,7} are all accounted for!
Another way to solve it or look at it is this:
C(5,0) + C(5,1) + C(5,2) + C(5,3) + C(5,4) + C(5,5)
1 + 5 + 10 + 10 + 5 + 1 = 32 ways
One way to have NONE of the elements with 1,2 and
Five ways to have ONE of the elements with 1,2 and
Ten ways to have TWO of the elements with 1,2 and
Ten ways to have THREE of the elements with 1,2 and
Five ways to have FOUR of the elements with 1,2 and
One way to have ALL FIVE of the elements {3,4,5,6,7} with 1,2.
---
Thirty-two ways...
g. subsets of A containing five elements, including 1, 2
{1, 2, _, _, _} with {3,4,5,6,7}
as the set of 5 choices for the 3 slots.
5*4*3
C(5,3) = ----- = 5*2 = 10 such subsets...
3*2*1
{1,2,3,4,5}, {1,2,3,5,6}, {1,2,4,5,6}, {1,2,5,6,7},
{1,2,3,4,6}, {1,2,3,5,7}, {1,2,4,5,7}, {1,2,3,4,7},
{1,2,3,6,7}, {1,2,4,6,7}
h. subsets of A with an even number of elements
C(7,0) 1
+ C(7,2) 21 = (7*6)/2
+ C(7,4) 35 = (7*6*5*4)/(4!)
+ C(7,6) 7
-------- ---
64
i. subsets of A with an odd number of elements
C(7,1) + C(7,3) + C(7,5) + C(7,7) =
7 + 35 + 21 + 1 = 64
Wow! Half of the 128 subsets have an ODD number of elements, and
half of hte 128 subsets have an EVEN cardinality.
Page 150, #4 a. i. TRUE
ii. FALSE 2 elementOf A but 2 NOT elementOf C (or E)
iii. FALSE 3 divides 9 but 6 does NOT divide 9 so
9 elementOf B but 9 NOT elementOf D
iv. TRUE
v. TRUE
vi. FALSE 4 elementOf complement of D, but
4 not elementOf complement of A
b. i. C intersection E is E
ii. B Union D is B
iii. A intersection B is D
iv. B intersection D is D
v. complement of A is Odd integers, or
{x | x = 2m + 1, m elementOf Z}
vi. A intersection E is E
Page 160, #6
a. If A and B are infinite sets, then A intersection B is infinite.
FALSE - Suppose A is the Even integers
and B is the Odd integers A intersect B is empty!
b. If B is infinite and A is subset of B, then A is infinite.
FALSE
Let B be the integers and A = { x | x > -1 and x < 10 }
A = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }
and A subsetOf B (aka Z).
c. If A subsetOf B with B finite, then A is finite.
TRUE
d. If A subsetOf B with A finite, then B is finite.
FALSE A = {1, 5, 7} B = Natural Numbers A subsetOf B
but B is infinite,
and |A| = 3
Page 161 #20 Professor Diane gave her chemistry class a test
consisting of three questions.
There are 21 students in her class, and every student
answered at least one question.
Five students did not answer the first question.
Seven failed to answer the second question.
Six did not answer the third question.
If nine students answered all three questions, how many
answered exactly one question?
-------------------------------------------------------
Let A represent answered Question 1
Let B represent answered Question 2
Let C represent answered Question 3
Let A U B represent the Union of sets A and B
Let AB represent the Intersection of sets A and B
Let cardinality of A be understood, so we need not
write | A | or cardinalityOf(A) here.
A = 21 - 5 = 16 students answered Q1
B = 21 - 7 = 14 students answered Q2
C = 21 - 6 = 15 students answered Q3
A U B U C = A + B + C - AB - AC - BC + ABC
21 = 16 + 14 + 15 - AB - AC - BC + 9
21 = 45 - (AB + AC + BC) + 9
21 = 54 - (AB + AC + BC)
(AB + AC + BC) = 54 - 21 = 33
AB = r + 9
AC = s + 9
BC = t + 9 AB+AC+BC = r + s + t + 27 = 33
r + s + t = 6
The ABC region of the Venn diagram is 9.
The sum of the r, s, t regions of the Venn diagram is 6.
r = AB - ABC
s = AC - ABC
t = BC - ABC
--- -- ---
6 = 33 - 27
Let w represent A - AB - AC + ABC or the upper leftmost
corner region of the Venn
diagram,
Let x represent B - AB - BC + ABC or the upper rightmost
corner region of the Venn
diagram,
Let y represent C - AC - BC + ABC or the bottommost
region of the Venn diagram.
w + x + y is what we want to know. w+x+y is the GOAL to find.
The sum of all the regions in the Venn diagram is 21.
21 = w + x + y + r + s + t + 9
21 = w + x + y + 6 + 9
21 = w + x + y + 15
w + x + y = 21 - 15 = 6 students answered EXACTLY one question.
DRAW the Venn diagram for this problem, and you will be able
to more easily follow the above argument, while you exercise
you skill with sets and the principle of inclusion/exclusion.
See Venn diagram and graphic GUI scanned solution
of the 3 question Chemistry Quiz problem.