Lecture 2. Properties of Combinations
June 25, 2014
Symmetry. We can say without calculations that 82 = 86 . Indeed, for every subset of
{1, 2, . . . , 8} of two elements there is a subset of six elements: its complement. For example,
{3, 5} corresponds to {1, 2, 4, 6, 7, 8}. This is a one-to-one correspondence.
So there are equally
many subsets of two elements and subsets of six elements. Similarly, 83 = 85 . More generally,
n
n
=
k
n−k
Power set. How many subsets does the set {1, 2, . . . , n} contain? Answer: 2n . Indeed, to
construct an arbitrary subset E, you should answer n questions:
• Is 1 ∈ E? Yes/No
• Is 2 ∈ E? Yes/No
• ...
• Is n ∈ E? Yes/No
For each question, there are two possible answers. The total number of choices is |2 · 2 ·{z. . . · 2} =
n times
2n . The set of all subsets of {1, . . . , n} is called a power set, and it contains 2n elements. But
we can also write the quantity of all subsets as the sum of binomial coefficients:
n
n
n
n
+
+
+ ... +
.
0
1
2
n
So we get the following identity:
n
n
n
+
+ ... +
= 2n
0
1
n
2
Example. n = 2. Then
∅, {1}, {2}, {1, 2}. Two ofthem
the set {1, 2} has 4 = 2 subsets:
2
2
have one element: 1 = 2, one has two elements, 2 = 1, and one has zero elements, 20 = 1.
Total: 1 + 2 + 1 = 4.
Reduction property. We can claim that
5
4
4
=
+
.
2
2
1
Indeed, the total number of subsets E ⊆ {1, 2, 3, 4, 5} which contain two elements is 52 . But
there are two possibilities:
Case 1. 5 ∈ E. Then E \ {5} is a one-element subset of {1, 2, 3, 4}; there are 41 such
subsets.
4
Case 2. 5 ∈
/ E. Then
E
is
a
two-element
subset
of
{1,
2,
3,
4}.
There
are
such subsets.
2
4
4
5
So 1 + 2 = 2 . In general,
n
n−1
n−1
=
+
k
k
k−1
Pascal’s Triangle.
n = 0:
n = 1:
1
n = 2:
1
n = 3:
n = 4:
1
1
4
1
0
2
1
3
6
4
2
0
3
0
1
n = 4:
4
0
2
2
3
1
n = 3:
2
1
n = 2:
1
1
1
n = 1:
1
3
0
0
n = 0:
1
3
2
4
1
3
3
4
2
4
3
Each element is the sum of two elements immediately
above it: this is the reduction formula.
n
n
We start from the edges, fill them with ones: 0 = n = 1, see the previous lecture. Then we
fill the inside from top to bottom using this rule, which is the reduction formula.
Newton’s Binomial Formula. We can expand (x + y)2 = x2 + 2xy + y 2 , and (x + y)3 =
3
x + 3x2 y + 3xy 2 + y 3 . The coefficients are taken from corresponding lines in Pascal’s triangle.
Why is this? Let us show this for n = 3.
(x + y)3 = (x + y)(x + y)(x + y) = xxx + xxy + xyx + yxx + xyy + yxy + yyx + yyy.
Each term has slots occupied by y: xxy ↔ {3},
yxy ↔ {1, 3}. If there is one slot occupied by
y, this corresponds to x2 y, and there are 31 such combinations. So we have: 31 x2 y. Other
terms give us:
3 3
3 2
3
3 3
2
x +
x y+
xy +
y .
0
1
2
3
The general formula looks like this:
n n
n n−1
n n
n
(x + y) =
x +
x y + ... +
y
0
1
n
Let x = y = 1. Then we get:
n
n
n
2 =
+
+ ... +
.
0
1
n
n
This formula was already proven above. Let x = 1, y = −1. Then
n
n
n
n
0=
−
+
−
+ ...,
0
1
2
3
so
n
n
n
n
+
+ ... =
+
+ ...
0
2
1
3
The quantity of subsets with even number of elements is equal to the quantity of subsets with
odd number of elements.
4
4