mat2348 : Discrete Mathematics
Mike Newman, 3 october 2021
assignment 1
[/20]
You should explain your reasoning. A correct “answer” with no explanation of where it came from is worth
very few marks, perhaps none.
As a general rule, you should try and simplify your expressions if it is reasonable to do so. But you may leave
your answers in the form of binomial coefficients, factorials, exponentials, etc.
1. Let A and B be finite sets.
[2]
a)
Show that (A × B) ∩ (B × A) = (A ∩ B) × (A ∩ B).
(Note: this is an equality of sets, so you should show that each element in (A × B) ∩ (B × A) is
in (A ∩ B) × (A ∩ B), and also that each element in (A ∩ B) × (A ∩ B) is in (A × B) ∩ (B × A).
[2]
b)
Using the previous part, or otherwise, show that |(A × B) ∪ (B × A)| = 2 |A| |B| − |A ∩ B| .
[2]
c)
Suppose that A = {1, 2, 3, 4, 5} and B = {4, 5, 6}. How many two-digit numbers can be formed
with one digit from A and the other digit from B?
2
Solution:
a)
Let z ∈ (A×B)∩(B ×A). Then z is an ordered pair (x, y) where (x, y) ∈ A×B and (x, y) ∈ B ×A.
This means that x ∈ A and y ∈ B, and also x ∈ B and y ∈ A. So x ∈ A and x ∈ B meaning
x ∈ A ∩ B; also y ∈ B and y ∈ A meaning y ∈ A ∩ B. Therefore z = (x, y) ∈ (A × B) ∩ (B × A).
Now let w ∈ (A ∩ B) × (A ∩ B). Then w is an ordered pair (f, g) where f ∈ A ∩ B and g ∈ A ∩ B.
This means that f ∈ A and g ∈ B meaning that (f, g) ∈ A × B, and also that f ∈ Bandg ∈ A
meaning that (f, g) ∈ N × A. Therefore w = (f, g) ∈ (A × B) ∩ (B × A).
b)
We start with the relationship between union and intersection, and apply the previous part.
|(A × B) ∪ (B × A)| = |A × B| + |A × B| − |(A × B) ∩ (B × A)|
= |A × B| + |B × A| − |(A ∩ B) × (B ∩ A)|
= |A| |B| + |B| |A| − |A ∩ B| |B ∩ A|
= |A| |B| + |A| |B| − |A ∩ B| |A ∩ B|
2
= 2 |A| |B| − |A ∩ B|
c)
2.
We want to know the size of (A × B) ∪ (B × A). We just derived a formula for this so the answer
is:
2
|(A × B) ∪ (B × A)| = 2 |A| |B| − |A ∩ B| = 2(5)(3) − (2)2 = 26
Consider the collection (multiset) of letters {{A, A, A, B, C, C, D, D, D, D, E}}.
[2]
a)
How many words can we make with these letters?
[2]
b)
How many of these words have all their letters in alphabetical order?
[2]
c)
How many of these words will contain both “CBC” and “DADADAD”?
In this context, a word is any sequence of all the given letters; the copies of one letter are indistinguishable.
Solution:
a)
This is the number of arrangements of a multiset; we saw this in class (Theorem 1.33).
11!
= 138600
(3!)(1!)(2!)(4!)(1!)
Of course the (1!) terms are redundant, we just include them to emphasize that we include all
the multiplicities.
b)
Only one. While we could permute the various A’s, this doesn’t change the arrangement. The
unique alphabetically sorted arrangement is AAABCCDDDDE.
c)
[3]
[2]
We could try this by an enormous number of cases, depending on where in the sequence we
have CAC and where we have DADADAD. Or we could think about it conceptually: we want
arrangments not of the multiset {{A, A, A, B, C, C, D, D, D, D, E}} containing eleven symbols, but
rather the multiset {{CBC, DADADAD, E}} containing three symbols. Some of these symbols
are bigger than the others, but that doesn’t matter. In fact this is a set containing three symbols.
So the number of arrangements is 3! = 6.
n
n n−1
.
3. Give two proofs of the fact that
=
k k−1
k
n
n−1
a) A combinatorial proof, by rewriting as
k=n
and interpreting each side as the result
k
k−1
of an application of the Product Principle. In particular, you should not use the formula for a
binomial coefficient, only its meaning in terms of counting subsets.
b)
An algebraic proof, by directly considering the formula for the binomial coefficients.
Solution:
a)
Let A = {1, 2, · · · , n}
Let C = {(C, p) : C ⊆ A , |C| = k , p ∈ C}; we will find |C|. There
are nk ways to choose C from
A, and there are k choices of p within C. So we get |C| = nk k.
Let S = {(p, S) : S ⊆ A , |S| = k − 1 , p ∈ A \ S}; we will find |S|. There are n ways to choose
p,
n−1
n−1
and then k−1 ways to choose S from the remaining elements of A. So we get |S| = n k−1 .
There is clearly a bijective correspondance between C and S. The ordered pair (C, p) ∈ C
corresponds to (p, C \ {p}) ∈ S, and the ordered pair (p, S) ∈ C corresponds to the ordered pair
(S ∪ {p} , p) ∈ C.
|C| = |S|
n
n−1
k=n
k
k−1
n n−1
n
=
k k−1
k
Comment: Notice that we didn’t need to know any formula for nk . All we need to know is that
it is a symolic shorthand for “the number of k-subsets chosen from an n-set”.
Comment: A nice way to think about this is to imagine having to choose a committee of k
individuals from a population of n individuals, and then choosing one of the committee memebrs
to be the president. Then C amounts to choosing the committee first and then choosing the
president among the committee members, and S amounts to choosing the president first from A
and then choosing the remaining committe members from A \ {p}.
b)
Since we have formulas for binomial coefficients, we can turn everything into factorials and happily
grind away!
n
n!
n
(n − 1)!
n
(n − 1)!
n n−1
=
= ·
= ·
=
k
k!(n − k)!
k (k − 1)!(n − k)!
k (k − 1)!((n − 1) − (k − 1))!
k k−1
Actually, that wasn’t so bad. On the other hand, the previous proof tells us this is not some
lucky coincidence of the formulas working out to be the same number, but rather that there is a
natural reason for them to be the same.
Solution:
a)
As a guide, 1mark for each way of computing and 1 way for equating. They don’t need to justify
the bijection.
In fact I expect most of them will answer along the lines of “Choose the committee
in nk ways then the president is k; or else choose the president in n ways then the rest of the
committee in n−1
k−1 . These are equal.” This is fine (I did a similar example in class).
b)
They know the formula for the binomial coefficient, so this should be straightforward. I imagine
most will get 2marks here.
[3]
n
4. Using the binomial theorem, prove that (r + 1) =
n X
n
j=0
Pn
n
j
rj for any r 6= 0.
Solution: Recall that (x + y)n = j=0 j xj y n−j . We set y = 1 to make all the powers of y go
away (they are all equal to 1), and set x = r.
n X
n j n−j
(x + y)n =
x y
j
j=0
n X
n
n
(r + 1) =
(r)j (1)n−j
j
j=0
n X
n j
n
(r + 1) =
r
j
j=0