Faculty of Sciences
Exam Sets and Combinatorics
Vrije Universiteit Amsterdam
Thursday, February 3, 2022 (18:45-21:00)
This exam consists of 5 questions and a total of 90 points can be obtained. The grade is calculated as
(number of points + 10)/10. Use of a calculator is not allowed.
Please provide an argument at every question!
1. In this exercise we consider the set-theoretic equality
(A\B)\(A ∩ C) = A\(B ∪ C).
(a) [7 points] Use Venn-diagrams to show that this equality holds for arbitrary sets A, B and C.
Solution: Draw Venn diagrams.
Grading: Left hand side: 1 point for drawing A\B, 1 point for drawing A ∩ C, and 1 point for
drawing (A\B)\(A ∩ C). Right hand side: 1 point for drawing A, 1 point for drawing B ∪ C, and 1
point for drawing A\(B ∪ C). 1 point for the conclusion that the two sides are the same.
(b) [9 points] Now prove the above equality, either by using the algebra of sets, or by formal reasoning
(your choice).
Solution:
• Using algebra of sets:
(A\B)\(A ∩ C)
Def of \
=
(A\B) ∩ (A ∩ C)c
de Morgan
=
=
(A ∩ B c ) ∩ (Ac ∪ C c )
Associativity
=
Identity law
=
Def of \
(A ∩ B c ) ∩ (A ∩ C)c
Distributivity
=
((A ∩ B c ) ∩ Ac ) ∪ ((A ∩ B c ) ∩ C c )
((A ∩ Ac ) ∩ B c ) ∪ (A ∩ (B c ∩ C c ))
(A ∩ (B c ∩ C c ))
de Morgan
=
complement law
A ∩ (B ∪ C)c
=
Def of \
=
(∅ ∩ B C ) ∪ (A ∩ (B c ∩ C c ))
A\(B ∪ C)
Grading: Each equation is 1 point. If all the names are missed then -2 points, if some, then -1
point.
• By formal reasoning: x ∈ (A\B)\(A ∩ C) ⇔ (x is in A\B) and (x is not in A ∩ C)
⇔ (x is in A and x is not in B) and (x is in (A ∩ C)c )
⇔ (x is in A and x is in B c ) and (x is in Ac ∪ C c )
⇔ (x ∈ A and x ∈ B c ) and (x ∈ Ac or x ∈ C c )
⇔ (x ∈ A and x ∈ B c and x ∈ Ac ) or (x ∈ A and x ∈ B c and x ∈ C c )
⇔ (x ∈ A and x ∈ B c and x ∈ C c ) ⇔ x ∈ A and x ∈ B c ∩ C c
⇔ x ∈ A and x ∈ (B ∪ C)c ⇔ x ∈ A ∩ (B ∪ C)c ⇔ x ∈ A\(B ∪ C).
2. In the universe U := [1, 3] consider the sets Ak (k ∈ N) given by
1 5
1
Ak := 1 + , − k .
k 2 2
1
(a) [5 points] Write Ack as a union of two intervals.
Solution:
Ack
1
5
1
= [1, 3]\Ak = 1, 1 +
∪
− ,3 .
k
2 2k
Grading: 0.5 points for each boundary point and each bracket. 1 point for taking the union of them
and writing down the whole expression.
(b) [5 points] Is the sequence of sets A1 , A2 , A3 , . . . an increasing or decreasing sequence?
Note: A sequence of sets A1 , A2 , . . . is an increasing if A1 ⊂ A2 ⊂ A3 ⊂ . . . and decreasing
of A1 ⊃ A2 ⊃ A3 ⊃ . . . .
Solution: Note that the left boundary points are decreasing, while the right boundary points are
increasing in k. [3 points] Therefore the intervals are getting larger as k increases so the sequence of
sets A1 , A2 , ... is increasing (A1 ⊂ A2 ⊂ A3 ⊂ ...). [2 points]
S∞
(c) [7 points] Determine a simple explicit expression for the set k=1 Ak .
Since
their unionSis the limit of the sets (as for each n ∈ N:
Sn the sequences of sets are increasing
S∞
n
A
=
A
hence
by
definition
A
n
k=1 k
k=1 k = limn→∞
k=1 Ak = limn→∞ An ). [3 points - the formal
1
argument is not required] Then note that limn→∞ (1 + n ) = 1 and limn→∞ ( 52 − 21n ) = 52 [2 points],
hence
1 5
1
5
lim An = lim 1 + , − n = 1,
,
n→∞
n→∞
n 2 2
2
as 1 and
5
2
are not included in any of the intervals Ak [2 points].
S∞
T∞
c
(d) [5 points] Determine a simple explicit expression for the set k=1 Ack = ( k=1 Ak ) .
T∞
S∞
c
Since
have
k=1 Ak = A1 [2 points]. Thus
k=1 Ak =
T∞ {Ak }c is anc increasing sequence of sets, we
1
5
c
( k=1 Ak ) = A1 [1 point]. Since A1 = 1 + 1, 2 − 2 = {2} [1 point] we have A1 = [1, 2) ∪ (2, 3] [1
point].
3. In a game chess between players A and B, there are three possible outcomes: a win for A (which
is a loss for B), a draw (tie), and a loss for A (which is a win for B).
Suppose players A and B plays 12 games.
(a) [7 points] What is the total number of possible outcomes?
This can be modelled by drawing balls with replacement with order [2 points] from an urn containing
three different outcomes for a chess game, thus with 3 different balls [2 points]. Since the players play
12 games, the total number of cases is 312 [3 points]. Alternatively: for each coordinate of the ordered
sequence of the three possible outcomes [2 points] we can choose altogether from 3 symbols [2 points].
Hence the total number of possible sequences are 3 · 3 · ... · 3 = 312 . [3 points]
(b) [8 points] How many outcomes are there in the event that player A does not win the first five
games and has three overall wins?
Since player A does not win the first 5 games, there are 2 different outcomes for each of these games
[2 points] so altogether there are 25 different possibilities for the first five
[2 points]. For the
games
7!
different possibilities
remaining seven games, player A has to win two games [1 point], hence 72 = 2!5!
[2 points]. Altogether there are 25 72 different possible outcomes [1 point].
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(c) [8 points] How many outcomes for the individual games are there, such that player A does not win
the first two games but overall player A wins exactly 5 games, player B wins at least 2 games and
there are at least 3 draws?
Since player A does not win the first two games, there are 22 possible outcomes for those games [1
point]. For the remaining 10 games, if player A wins 5 of them, player B wins 2 of them and the
remaining 3 are a draw than altogether player B wins at least 2 games and there are at least 3 draws
10
10!
[2 points]. This can can be computed by the multinomial coefficient 5,2,3
= 5!2!3!
[4 points]. Hence
2 10
there are 2 5,2,3 different possible outcomes [1 point].
4. [15 points] Using mathematical induction, prove that
n
X
k=1
n
1
=
k(k + 1)
n+1
for all n ∈ N.
Base case: for n = 1 we have that
1
X
k=1
Inductive step: Assume that
n+1
n+2 also holds [3 points].
n+1
X
k=1
1
1
1
1
=
= =
k(k + 1)
1·2
2
1+1
Pn
1
k=1 k(k+1)
=
n
n+1
[5 points].
holds (Inductive Hypothesis) and prove that
Pn+1
1
k=1 k(k+1)
n
X
1
1
1
=
+
k(k + 1)
k(k + 1) (n + 1)(n + 2)
[2 points]
k=1
1
n
+
[2 points]
n + 1 (n + 1)(n + 2)
n(n + 2) + 1
(n + 1)2
n+1
=
=
=
,
(n + 1)(n + 2)
(n + 1)(n + 2)
n+2
IH
=
as required. [2 points]
5. Let f : [ − π, π]2 → R2 be the function given by
f (x, z) := (sin x, 2 cos z)
Consider the sets A := (− π2 , π2 ]2 and B := [0, 1) × (−2, 1].
(a) [7 points] Determine the image of A under f .
The image of the set A = (− π2 , π2 ] under the function f1 (x) = sin x is (−1, 1] [2 points]. The image of
the set (− π2 , π2 ] under the function cos z is [0, 1] (as cos π2 = 0 the limit at 0 is included) and thus for
the function f2 (z) = 2 cos z, it is [0, 2] [2 points]. Since f (x, z) = (f1 (x), f2 (z)) the image of (− π2 , π2 ]2
under f is (−1, 1] × [0, 2]. [3 points]
(b) [7 points] Determine the inverse image of B under f .
3
=
The inverse image f −1 (B) of the set B = [0, 1) × (−2, 1] under the function f is given by f1−1 ([0, 1)) ×
f2−1 ((−2, 1]) ⊂ [−π, π]2 where f1 = sin x and f2 = 2 cos x [2 points]. Since the sine function has image
[0, 1) for the interval [0, π), we have f1−1 ([0, 1)) = [0, π) [1 point]. The function f2 = 2 cos x has image
(−2, 1] when cos x has image (−1, 12 ] [1 point]. Since cos x = −1 implies x = π or x = −π and cos x = 21
implies x = π3 or x = − π3 in the domain [−π, π], we have cos−1 ((−1, 12 ]) = (−π, − π3 ] ∪ [ π3 , π) [1 point].
Hence f2−1 ((−2, 1]) = (−π, − π3 ]∪[ π3 , π) [1 point]. Hence f −1 (B) = [0, 1) × (−π, − π3 ] ∪ [0, 1) × [ π3 , π)
[1 point].
4