Binomial Coefficients and Generating Functions - Cs.ioc.ee

Binomial Coefficients and Generating Functions
ITT9130 Konkreetne Matemaatika
Chapter Five
Basic Identities
Basic Practice
Tricks of the Trade
Generating Functions
Hypergeometric Functions
Hypergeometric Transformations
Partial Hypergeometric Sums

Contents
1 Binomial coefficients
2 Generating Functions
Operations on Generating Functions
Building Generating Functions that Count
Identities in Pascal’s Triangle

Next section
1 Binomial coefficients
2 Generating Functions
Operations on Generating Functions
Building Generating Functions that Count
Identities in Pascal’s Triangle

Binomial theorem
Theorem 1
for any integer n 0.
(a + b) n =
n
∑
k=0
n
a
k
k b n−k ,
Proof. Expanding (a + b) n = (a + b)(a + b)···(a + b) yields the sum of the
2 n products of the form e1e2 ···en, where each ei is a or b. These
terms are composed by selecting from each factor (a + b) either a or
b. For example, if we select a k times, then we must choose b n − k
times. So, we can rearrange the sum as
(a + b) n n
= ∑ Cka
k=0
k b n−k ,
where the coefficient Ck is the number how many possibilities are to
select k elements (k factors (a + b)) from a set of n elements (from
the production of n factors (a + b)(a + b)···(a + b)).
That is why the coefficient Ck is called "(from) n choose k" and
denoted by n k . Q.E.D.

Binomial coefficients and combinations
Theorem 2
The number of k-subsets of an n-set is
n n!
=
k k!(n − k)!
Proof. At first, determine the number of k-element sequences: there are n
choices for the first element of the sequence; for each, there are n − 1
choices for the second; and so on, until there are n − k + 1 choices
for the k-th. This gives n(n − 1)...(n − k + 1) = n k choices in all. And
since each k-element subset has exactly k! different orderings, this
number of sequences counts each subset exactly k! times. To get
our answer, divide by k!:
n
k
= nk
k! =
n!
k!(n − k)!
Some other notations used for the "n choose k" in literature:
C n k ,C(n,k), nCk, n Ck.
Q.E.D.

Binomial coefficients and combinations
Theorem 2
The number of k-subsets of an n-set is
n n!
=
k k!(n − k)!
Proof. At first, determine the number of k-element sequences: there are n
choices for the first element of the sequence; for each, there are n − 1
choices for the second; and so on, until there are n − k + 1 choices
for the k-th. This gives n(n − 1)...(n − k + 1) = n k choices in all. And
since each k-element subset has exactly k! different orderings, this
number of sequences counts each subset exactly k! times. To get
our answer, divide by k!:
n
k
= nk
k! =
n!
k!(n − k)!
Some other notations used for the "n choose k" in literature:
C n k ,C(n,k), nCk, n Ck.
Q.E.D.

Properties of Binomial Coefficients
n
k = 2n (number of all subsets of n-set);
= 0 (number of subsets with odd cardinality is
equel to the number of sets with even cardinality).
1 ∑ n k=0
2 ∑ n k=0 (−1)k n
k
Tõestus. Take a = b = 1 in the binomial theorem:
n
n n
=
k ∑ k
∑
k=0
n
∑
k=0
k=0
Take a = −1 ja b = 1:
n
(−1) k
n n
=
k ∑
k=0
n
k
1 k 1 n−k = (1 + 1) n = 2 n
(−1) k 1 n−k = (−1 + 1) n = 0
Q.E.D.

Another Property
3
n n
k = n−k ;
Proof. Direct conclusion from the Theorem 2:
n
k
=
n!
k!(n − k)! =
n
n − k
Q.E.D.

Yet Another Proerty
4 If n,k > 0, then n−1
k−1
Proof.
Note that
+ n−1
k
n
= k .
1 1 k + n − k
+ =
n − k k k(n − k) =
n
k(n − k) .
Multiplying this by (n − 1)! and dividing by (k − 1)!(n − k − 1)!, we get
(n − 1)!
(k − 1)!(n − k)(n − k − 1)! +
(n − 1)!
k(k − 1)!(n − k − 1)! =
n(n − 1)!
k(k − 1)!(n − k)(n − k − 1)!
This can be rewritten after some simplifying transformations as:
(n − 1)!
(k − 1)!(n − k)! +
(n − 1)!
k!(n − k − 1)! =
n!
k!(n − k)!
Q.E.D.

Pascal’s Triangle
n
0
n
0
1
n
1
n
2
n
3
n
4
n
5
n
6
1 1 1
2 1 2 1
3 1 3 3 1
4 1 4 6 4 1
5 1 5 10 10 5 1
6 1 6 15 20 15 6 1
Blaise Pascal
(1623–1662)

Pascal’s Triangle
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
Pascal Triangle is symmetric with respect to the vertical line
through its apex.
Blaise Pascal
(1623–1662)
Every number is the sum of the two numbers immediately above it.

Pascal’s Triangle
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
Pascal Triangle is symmetric with respect to the vertical line
through its apex.
Blaise Pascal
(1623–1662)
Every number is the sum of the two numbers immediately above it.

Next section
1 Binomial coefficients
2 Generating Functions
Operations on Generating Functions
Building Generating Functions that Count
Identities in Pascal’s Triangle

Generating Functions
Definition
The generating function for the infinite sequence 〈g0,g1,g2,...〉 is
the power series
G(x) = g0 + g1x + g2x 2 + g3x 3 + ··· =
Ssome simple examples
∞
∑
n=0
〈0,0,0,0,...〉 ←→ 0 + 0x + 0x 2 + 0x 3 + ··· = 0
gnx n
〈1,0,0,0,...〉 ←→ 1 + 0x + 0x 2 + 0x 3 + ··· = 1
〈2,3,1,0,...〉 ←→ 2 + 3x + 1x 2 + 0x 3 + ··· = 2 + 3x + 1x 2

More examples (1)
〈1,1,1,1,...〉 ←→ 1 + x + x 2 + x 3 + ··· = 1
1−x
Subtract the equations:
S = 1 + x + x 2 + x 3 + ···
xS = x + x 2 + x 3 + ···
(1 − x)S = 1 ehk S = 1
1 − x
NB! This formula converges only for −1 < x < 1.
Actually, we don’t worry about convergence issues.

More examples (1)
〈1,1,1,1,...〉 ←→ 1 + x + x 2 + x 3 + ··· = 1
1−x
Subtract the equations:
S = 1 + x + x 2 + x 3 + ···
xS = x + x 2 + x 3 + ···
(1 − x)S = 1 ehk S = 1
1 − x
NB! This formula converges only for −1 < x < 1.
Actually, we don’t worry about convergence issues.

More examples (2)
〈a,ab,ab 2 ,ab 3 ,...〉 ←→ a + abx + ab 2 x 2 + ab 3 x 3 + ··· = a
1−bx
Like in the previous example:
Subtract and get:
S = a + abx + ab 2 x 2 + ab 3 x 3 + ···
bxS = abx + ab 2 x 2 + ab 3 x 3 + ···
(1 − bx)S = a ehk S = a
1 − bx

More examples (3)
Taking in the last example a = 0,5 and b = 1 yields
0,5 + 0,5x + 0,5x 2 + 0,5x 3 + ··· = 0,5
1 − x
Taking a = 0,5 and b = −1, gives
0,5 − 0,5x + 0,5x 2 − 0,5x 3 + ··· = 0,5
1 + x
Adding equations (1) and (2), we get the generating function of the
sequence 〈1,0,1,0,1,0,...〉:
1 + x 2 + x 4 + x 6 + ··· = 0,5 0,5
+
1 − x 1 + x =
1
1
=
(1 − x)(1 + x) 1 − x 2
(1)
(2)

More examples (3)
Taking in the last example a = 0,5 and b = 1 yields
0,5 + 0,5x + 0,5x 2 + 0,5x 3 + ··· = 0,5
1 − x
Taking a = 0,5 and b = −1, gives
0,5 − 0,5x + 0,5x 2 − 0,5x 3 + ··· = 0,5
1 + x
Adding equations (1) and (2), we get the generating function of the
sequence 〈1,0,1,0,1,0,...〉:
1 + x 2 + x 4 + x 6 + ··· = 0,5 0,5
+
1 − x 1 + x =
1
1
=
(1 − x)(1 + x) 1 − x 2
(1)
(2)

More examples (3)
Taking in the last example a = 0,5 and b = 1 yields
0,5 + 0,5x + 0,5x 2 + 0,5x 3 + ··· = 0,5
1 − x
Taking a = 0,5 and b = −1, gives
0,5 − 0,5x + 0,5x 2 − 0,5x 3 + ··· = 0,5
1 + x
Adding equations (1) and (2), we get the generating function of the
sequence 〈1,0,1,0,1,0,...〉:
1 + x 2 + x 4 + x 6 + ··· = 0,5 0,5
+
1 − x 1 + x =
1
1
=
(1 − x)(1 + x) 1 − x 2
(1)
(2)

Next subsection
1 Binomial coefficients
2 Generating Functions
Operations on Generating Functions
Building Generating Functions that Count
Identities in Pascal’s Triangle

Operations on Generating Functions
1. Scaling
If
then
for any c ∈ R.
Proof.
〈f0,f1,f2,...〉 ←→ F (x),
〈cf0,cf1,cf2,...〉 ←→ c · F (x),
〈cf0,cf1,cf2,...〉 ←→ cf0 + cf1x + cf2x 2 + ··· =
= c(f0 + f1x + f2x 2 + ···) =
= cF (x)
Q.E.D.

Operations on Generating Functions (2)
2. Addition
If 〈f0,f1,f2,...〉 ←→ F (x) and 〈g0,g1,g2,...〉 ←→ G(x), then
Proof.
〈f0 + g0,f1 + g1,f2 + g2,...〉 ←→ F (x) + G(x).
〈f0 + g0, f1 + g1, f2 + g2, ...〉 ←→
=
∞
∑
n=0
∞
∑ fnx
n=0
n
∞
+ ∑ gnx
n=0
n
= F (x) + G(x)
(fn + gn)x n =
=
Q.E.D.

Operations on Generating Functions (3)
3. Right-shift
If 〈f0,f1,f2,...〉 ←→ F (x), then
Proof.
〈0,0,...,0 ,f0,f1,f2,...〉 ←→ x k · F (x).
k zeros
〈0,0,...,0,f0,f1,f2,...〉 ←→ f0x k + f1x k+1 + f2x k+2 + ... =
= x k (f0 + f1x + f2x 2 + ···) =
= x k · F (x)
Q.E.D.

Operations on Generating Functions (4)
4. Differentiation
If 〈f0,f1,f2,...〉 ←→ F (x), then
Proof.
〈f1,2f2,3f3,...〉 ←→ F ′ (x).
〈f1,2f2,3f3,...〉 ←→ f1 + 2f2x + 3f3x 2 + ... =
= d
dx (f0 + f1x + f2x 2 + f3x 3 + ···) =
= d
F (x)
dx
Q.E.D.

Operations on Generating Functions (4)
4. Differentiation
If 〈f0,f1,f2,...〉 ←→ F (x), then
Example
〈1,1,1,...〉 ←→ 1
1−x
〈1,2,3,...〉 ←→ d
dx
〈f1,2f2,3f3,...〉 ←→ F ′ (x).
1 1
1−x =
(1−x) 2

Operations on Generating Functions (5)
5. Integration
If 〈f0,f1,f2,...〉 ←→ F (x), then
Proof.
〈0,f0, 1 1 1
f1, f2,
2 3 4 f3,...〉 ←→
x
0
F (z)dz.
〈0,f0, 1 1 1
f1, f2,
2 3 4 f3,...〉 ←→ f0x + 1
2 f1x 2 + 1
3 f2x 3 + 1
4 f3x 4 + ... =
=
=
x
(f0 + f1z + f2z
0
2 + f3z 3 + ···)dz =
x
0
F (z)dz
Q.E.D.

Operations on Generating Functions (5)
5. Integration
If 〈f0,f1,f2,...〉 ←→ F (x), then
Example
〈0,f0, 1 1 1
f1, f2,
2 3 4 f3,...〉 ←→
〈1,1,1,1...〉 ←→ 1
1−x
〈0,1, 1 1
2 , 3 ,...〉 ←→ x
0
x
0
F (z)dz.
1
1
1−z dz = −ln(1 − x) = ln (1−x)

Operations on Generating Functions (6)
6. Convolution (product)
If 〈f0,f1,f2,...〉 ←→ F (x) and 〈g0,g1,g2,...〉 ←→ G(x), then
〈h0,h1,h2,...〉 ←→ F (x) · G(x),
where hn = f0gn + f1gn−1 + f2gn−2 + ··· + fng0.

Operations on Generating Functions (6)
6. Convolution (product)
If 〈f0,f1,f2,...〉 ←→ F (x) and 〈g0,g1,g2,...〉 ←→ G(x), then
〈h0,h1,h2,...〉 ←→ F (x) · G(x),
where hn = f0gn + f1gn−1 + f2gn−2 + ··· + fng0.
Proof.
F (x) · G(x) = (f0 + f1x + f2x 2 + ...)(g0 + g1x + g2x 2 + ...)
= f0g0 + (f0g1 + f1g0)x + (f0g2 + f1g1 + f2g0)x 2 + ...
Q.E.D.

Operations on Generating Functions (6)
6. Convolution (product)
If 〈f0,f1,f2,...〉 ←→ F (x) and 〈g0,g1,g2,...〉 ←→ G(x), then
〈h0,h1,h2,...〉 ←→ F (x) · G(x),
where hn = f0gn + f1gn−1 + f2gn−2 + ··· + fng0.
Proof.
F (x) · G(x) = (f0 + f1x + f2x 2 + ...)(g0 + g1x + g2x 2 + ...)
= f0g0 + (f0g1 + f1g0)x + (f0g2 + f1g1 + f2g0)x 2 + ...
Notice that all terms involving the same power of xlie on a /sloped diagonal:
g0x 0 g1x 1 g2x 2 g3x 3 a0x
...
0 f0g0x 0 f0g1x 1 f0g2x 2 f0g3x 3 f1x
...
1 f1g0x 1 f1g1x 2 f1g2x 3 f2x
...
2 f2g0x 2 f2x 3 f3x
...
3 f3g0x 3 ...
. ...
Q.E.D.

Operations on Generating Functions (6)
6. Convolution (product)
If 〈f0,f1,f2,...〉 ←→ F (x) and 〈g0,g1,g2,...〉 ←→ G(x), then
〈h0,h1,h2,...〉 ←→ F (x) · G(x),
where hn = f0gn + f1gn−1 + f2gn−2 + ··· + fng0.
Example
Hence
〈1,1,1,1...〉 · 〈0,1, 1 1
, ,...〉 =
2 3
= 〈1·0, 1·0 + 1·1, 1·0 + 1·1 + 1· 1
2
= 〈0, 1, 1 + 1 1 1
, 1 + +
2 2 3 ,···〉
= 〈0, H1, H2, H3,···〉
∑ Hkx
k0
k =
1
(1 − x) ln
1
(1 − x) .
1 1
, 1·0 + 1·1 + 1· + 1·
2 3 ,···〉

Operations on Generating Functions (7)
Example
〈1,1,1,1,...〉 ←→ 1
1 − x
〈1,2,3,4,...〉 ←→ d 1
dx 1 − x =
1
(1 − x) 2
1 x
〈0,1,2,3,...〉 ←→ x · =
(1 − x) 2 (1 − x) 2
〈1,4,9,16,...〉 ←→ d x 1 + x
=
dx (1 − x) 2 (1 − x) 3
1 + x x(1 + x)
〈0,1,4,9,...〉 ←→ x · =
(1 − x) 3 (1 − x) 3

Counting with Generating Functions
Choosing k-subset from n-set
Binomial theorem yields:
n n n n
, , ,..., ,0,0,0,... ←→
0 1 2 n
n n n
←→ + x + x
0 1 2
2
n
...,
n
x n = (1 + x) n
Thus, the coefficient of x k in (1 + x) n is the number of ways to
choose k distinct items from a set of size n.
For example, the coefficient of x 2 is , the number of ways to choose
2 items from a set with n elements.
Similarly, the coefficient of x n+1 is the number of ways to choose
n + 1 items from a n-set, which is zero.

Counting with Generating Functions
Choosing k-subset from n-set
Binomial theorem yields:
n n n n
, , ,..., ,0,0,0,... ←→
0 1 2 n
n n n
←→ + x + x
0 1 2
2
n
...,
n
x n = (1 + x) n
Thus, the coefficient of x k in (1 + x) n is the number of ways to
choose k distinct items from a set of size n.
For example, the coefficient of x 2 is , the number of ways to choose
2 items from a set with n elements.
Similarly, the coefficient of x n+1 is the number of ways to choose
n + 1 items from a n-set, which is zero.

Next subsection
1 Binomial coefficients
2 Generating Functions
Operations on Generating Functions
Building Generating Functions that Count
Identities in Pascal’s Triangle

Building Generating Functions that Count
.
The generating function for the number of ways to choose n elements from this
set A is the function A(x) that’s expansion into power series has coefficient
ai = 1 of x i iff i can be selected into the subset, otherwise ai = 0
Examples of GF selecting items from a set A :
If any natural number can be selected:
A(x) = 1 + x + x 2 + x 3 + ··· = 1
1 − x

Building Generating Functions that Count
.
The generating function for the number of ways to choose n elements from this
set A is the function A(x) that’s expansion into power series has coefficient
ai = 1 of x i iff i can be selected into the subset, otherwise ai = 0
Examples of GF selecting items from a set A :
If any natural number can be selected:
If any even number can be selected:
A(x) = 1 + x + x 2 + x 3 + ··· = 1
1 − x
A(x) = 1 + x 2 + x 4 + x 6 + ··· = 1
1 − x 2

Building Generating Functions that Count
.
The generating function for the number of ways to choose n elements from this
set A is the function A(x) that’s expansion into power series has coefficient
ai = 1 of x i iff i can be selected into the subset, otherwise ai = 0
Examples of GF selecting items from a set A :
If any natural number can be selected:
If any even number can be selected:
A(x) = 1 + x + x 2 + x 3 + ··· = 1
1 − x
A(x) = 1 + x 2 + x 4 + x 6 + ··· = 1
1 − x 2
If any positive even number can be selected:
A(x) = x 2 + x 4 + x 6 + ··· = x
1 − x 2

Building Generating Functions that Count
.
The generating function for the number of ways to choose n elements from this
set A is the function A(x) that’s expansion into power series has coefficient
ai = 1 of x i iff i can be selected into the subset, otherwise ai = 0
Examples of GF selecting items from a set A :
If any natural number can be selected:
A(x) = 1 + x + x 2 + x 3 + ··· = 1
1 − x
If any number multiple of 5 can be selected:
A(x) = 1 + x 5 + x 10 + x 15 + ··· = 1
1 − x 5

Building Generating Functions that Count
.
The generating function for the number of ways to choose n elements from this
set A is the function A(x) that’s expansion into power series has coefficient
ai = 1 of x i iff i can be selected into the subset, otherwise ai = 0
Examples of GF selecting items from a set A :
If any natural number can be selected:
If at most four can be selected:
A(x) = 1 + x + x 2 + x 3 + ··· = 1
1 − x
A(x) = 1 + x + x 2 + x 3 + x 4 =
If zero or one number can be selected:
A(x) = 1 + x
1 − x5
1 − x

Counting elements of two sets
Convolution Rule
Let A(x) be the generating function for selecting items from set A , and
let B(x), be the generating function for selecting items from set B.
If A and B are disjoint, then the generating function for selecting items from
the union A ∪ B is the product A(x) · B(x).
Proof. To count the number of ways to select n items from A ∪ B we have to select j
items from A and n − j items from B, where j ∈ {0,1,2,...,n}.
Summing over all the possible values of j gives a total of
a0bn + a1bn−1 + a2bn−2 + ··· + anb0
ways to select n items from A ∪ B. This is precisely the coefficient of x n in the series
for A(x) · B(x) Q.E.D.

How many positive integer solutions does the equation
x1 + x2 = n have?
We accept any natural number can be solution for x1, i.e generating function for
selection a value for this variable is A(x) = 1 + x + x2 + x3 + ··· = 1
1−x ;
same for variable x2;
all pairs of numbers for (x1,x2) are selected by the function H(x) = A(x) · A(x);
numer of solutions is the coefficient of xn .
H(x) = (1 + x + x 2 + x 3 + ···)(1 + x + x 2 + x 3 + ···) =
= 1 · (1 + x + x 2 + x 3 + ···) + x(1 + x + x 2 + x 3 + ···) +
+x 2 (1 + x + x 2 + x 3 + ···) + x 3 (1 + x + x 2 + x 3 + ···) + ··· =
= 1 + 2x + 3x 2 + 4x 3 + ··· + (n + 1)x n 1
+ ··· =
(1 − x) 2
Indeed, this equation has n + 1 solutions:
0 + n = n
1 + (n − 1) = n
2 + (n − 2) = n
··· ···
n + 0 = n

How many positive integer solutions does the equation
x1 + x2 = n have?
We accept any natural number can be solution for x1, i.e generating function for
selection a value for this variable is A(x) = 1 + x + x2 + x3 + ··· = 1
1−x ;
same for variable x2;
all pairs of numbers for (x1,x2) are selected by the function H(x) = A(x) · A(x);
numer of solutions is the coefficient of xn .
H(x) = (1 + x + x 2 + x 3 + ···)(1 + x + x 2 + x 3 + ···) =
= 1 · (1 + x + x 2 + x 3 + ···) + x(1 + x + x 2 + x 3 + ···) +
+x 2 (1 + x + x 2 + x 3 + ···) + x 3 (1 + x + x 2 + x 3 + ···) + ··· =
= 1 + 2x + 3x 2 + 4x 3 + ··· + (n + 1)x n 1
+ ··· =
(1 − x) 2
Indeed, this equation has n + 1 solutions:
0 + n = n
1 + (n − 1) = n
2 + (n − 2) = n
··· ···
n + 0 = n

How many positive integer solutions does the equation
x1 + x2 = n have?
We accept any natural number can be solution for x1, i.e generating function for
selection a value for this variable is A(x) = 1 + x + x2 + x3 + ··· = 1
1−x ;
same for variable x2;
all pairs of numbers for (x1,x2) are selected by the function H(x) = A(x) · A(x);
numer of solutions is the coefficient of xn .
H(x) = (1 + x + x 2 + x 3 + ···)(1 + x + x 2 + x 3 + ···) =
= 1 · (1 + x + x 2 + x 3 + ···) + x(1 + x + x 2 + x 3 + ···) +
+x 2 (1 + x + x 2 + x 3 + ···) + x 3 (1 + x + x 2 + x 3 + ···) + ··· =
= 1 + 2x + 3x 2 + 4x 3 + ··· + (n + 1)x n 1
+ ··· =
(1 − x) 2
Indeed, this equation has n + 1 solutions:
0 + n = n
1 + (n − 1) = n
2 + (n − 2) = n
··· ···
n + 0 = n

Number of solutions of the equation x1 + x2 + ··· + xk = n
Theorem
The number of ways to distribute n identical objects
into k bins is n+k−1
n .
Proof.
The number of ways to distribute n objects equals to the number of solutions of
x1 + x2 + ··· + xk = n that is coefficient of x n of the generating function
G(x) = 1/(1 − x) k = (1 − x) −k .
For recollection: Maclaurin series (a Taylor series
expansion of a function about 0):
.
f (x) = f (0) + f ′ (0)x + f ′′ (0)
2! x2 + f ′′′ (0)
x
3!
3 + ··· + f (n) (0)
x
n!
n + ···

Number of solutions of the equation x1 + x2 + ··· + xk = n
Theorem
The number of ways to distribute n identical objects
into k bins is n+k−1
n .
Proof.
The number of ways to distribute n objects equals to the number of solutions of
x1 + x2 + ··· + xk = n that is coefficient of x n of the generating function
G(x) = 1/(1 − x) k = (1 − x) −k .
For recollection: Maclaurin series (a Taylor series
expansion of a function about 0):
.
f (x) = f (0) + f ′ (0)x + f ′′ (0)
2! x2 + f ′′′ (0)
x
3!
3 + ··· + f (n) (0)
x
n!
n + ···

Equation x1 + x2 + ··· + xk = n (2)
Continuation of the proof ...
let’s differentiate G(x) = (1 − x) −k :
G ′ (x) = k(1 − x) −(k+1)
G ′′ (x) = k(k + 1)(1 − x) −(k+2)
G ′′′ (x) = k(k + 1)(k + 2)(1 − x) −(k+3)
G (n) (x) = k(k + 1)···(k + n − 1)(1 − x) −(k+n)
Coefficient of x n can be evaluated as:
G (n) (0)
n!
= k(k + 1)···(k + n − 1)
= (k + n − 1)!
=
=
n!
(k − 1)!n! =
n + k − 1
n
Q.E.D.

Distribute n objects into k bins so that there is at least one
object in each bin
Theorem
The number of positive solutions of the equation
x1 + x2 + ··· + xk = n is n−1
k−1 .
Idea of the proof. Possible number of objects in a single bin (xi > 0)
could be generated by the function
C(x) = x + x 2 + x 3 + ··· = x(1 + x + x 2 + ···) = x
1 − x
Similarly to the previous theorem, the number distributions is the
coefficient of x n of the generating function
H(X ) = C k (x) =
k x
(1 − x) k
Q.E.D.

Example: 100 Euro
How many ways 100 Euro can be changed using smaller banknotes?
Generating functions for selecting banknotes of 5, 10, 20 or 50 Euros:
Generating function for the task
A(x) = x 0 + x 5 + x 10 + x 15 + ··· = 1
1 − x5 B(x) = x 0 + x 10 + x 20 + x 30 1
+ ··· =
1 − x10 C(x) = x 0 + x 20 + x 40 + x 60 1
+ ··· =
1 − x20 D(x) = x 0 + x 50 + x 100 + x 150 1
+ ··· =
1 − x50 P(x) = A(x)B(x)C(x)D(x) =
1
(1 − x 5 )(1 − x 10 )(1 − x 20 )(1 − x 50 )

Example: 100 Euro
How many ways 100 Euro can be changed using smaller banknotes?
Generating functions for selecting banknotes of 5, 10, 20 or 50 Euros:
Generating function for the task
A(x) = x 0 + x 5 + x 10 + x 15 + ··· = 1
1 − x5 B(x) = x 0 + x 10 + x 20 + x 30 1
+ ··· =
1 − x10 C(x) = x 0 + x 20 + x 40 + x 60 1
+ ··· =
1 − x20 D(x) = x 0 + x 50 + x 100 + x 150 1
+ ··· =
1 − x50 P(x) = A(x)B(x)C(x)D(x) =
1
(1 − x 5 )(1 − x 10 )(1 − x 20 )(1 − x 50 )

Example: 100 Euro
How many ways 100 Euro can be changed using smaller banknotes?
Generating functions for selecting banknotes of 5, 10, 20 or 50 Euros:
Generating function for the task
A(x) = x 0 + x 5 + x 10 + x 15 + ··· = 1
1 − x5 B(x) = x 0 + x 10 + x 20 + x 30 1
+ ··· =
1 − x10 C(x) = x 0 + x 20 + x 40 + x 60 1
+ ··· =
1 − x20 D(x) = x 0 + x 50 + x 100 + x 150 1
+ ··· =
1 − x50 P(x) = A(x)B(x)C(x)D(x) =
1
(1 − x 5 )(1 − x 10 )(1 − x 20 )(1 − x 50 )

Example: 100 Euro (2)
1. Observation:
(1 − x 5 )(1 + x 5 + ··· + x 45 + 2x 50 + 2x 55 + ··· + 2x 95 + 3x 100 + 3x 105 + ··· + 3x 145 + 4x 150 + ···) =
Thus:
1 + x 5 + ··· + x 45 + 2x 50 + 2x 55 + ··· + 2x 95 + 3x 100 + 3x 105 + ··· + 3x 145 + 4x 150 ··· −
− x 5 − ··· − x 45 − x 50 − 2x 55 − ··· − 2x 95 − 2x 100 − 3x 105 − ··· − 3x 145 − 3x 150 − 4x 155 − ··· =
= 1 + x 50 + x 100 + x 150 + x 200 1
+ ··· =
1 − x50 F (x) = A(x)D(x) =
1
(1 − x5 )(1 − x50 ) =
k
∑ + 1 x
k0 10
5k = ∑ fkx
k0
5k

Example: 100 Euro (2)
1. Observation:
(1 − x 5 )(1 + x 5 + ··· + x 45 + 2x 50 + 2x 55 + ··· + 2x 95 + 3x 100 + 3x 105 + ··· + 3x 145 + 4x 150 + ···) =
Thus:
1 + x 5 + ··· + x 45 + 2x 50 + 2x 55 + ··· + 2x 95 + 3x 100 + 3x 105 + ··· + 3x 145 + 4x 150 ··· −
− x 5 − ··· − x 45 − x 50 − 2x 55 − ··· − 2x 95 − 2x 100 − 3x 105 − ··· − 3x 145 − 3x 150 − 4x 155 − ··· =
= 1 + x 50 + x 100 + x 150 + x 200 1
+ ··· =
1 − x50 2. Similarly:
F (x) = A(x)D(x) =
G(x) = B(x)C(x) =
1
(1 − x5 )(1 − x50 ) =
k
∑ + 1 x
k0 10
5k = ∑ fkx
k0
5k
1
(1 − x10 )(1 − x20 ) =
k
∑ + 1 x
k0 2
10k = ∑ gkx
k0
10k

Example: 100 Euro (3)
Convolution:
The coefficient of x 100 equals to
P(x) = F (x)G(x) = ∑ cnx
k0
5n
c20 = f0g10 + f2g9 + f4g8 + ··· + f20g0
10
∑
= f2kg10−k
k=0
10
2k 10 − k
= ∑ + 1
+ 1
k=0 10
2
10
k + 5 12 − k
= ∑
k=0 5 2
= 1(6 + 5 + 5 + 4 + 4) + 2(3 + 3 + 2 + 2 + 1) + 3 · 1 =
= 24 + 22 + 3 = 49

Next subsection
1 Binomial coefficients
2 Generating Functions
Operations on Generating Functions
Building Generating Functions that Count
Identities in Pascal’s Triangle

Vandermonde’s identity
Binomial theorem provides the generating function:
Convolution yields
(1 + x) m
m
= ∑ x
k0 k
k
∑
k0
m + n
k
x k = (1 + x) m+n
and (1 + x) n
n
= ∑ x
k0 k
k
= (1 + x) m · (1 + x) n
∑
m
= x
k0 k
k
n
∑ x
k0 k
k
k
m n
= ∑ ∑
k0 j=0 j k − j
x k

Vandermonde’s identity (2)
Equating coefficients of z k on both sides of this equation gives:
Convolution yields
1
1
1 9
1
8
1
7
1
6
28
1
5
21
36 84
1
4
15
56
1
3
10
35
1
2
6
20
70
Vandermonde’s convolution:
k
m n m + n
∑
=
j=0 j k − j k
1
3
10
35
126 126
1
4
15
56
1
5
21
1
6
28
1
7
84 36
1
8
1
1
9 1
9 3 6 3 6 3 6
= + +
2 0 2 1 1 2 0

Vandermonde’s identity (2)
Equating coefficients of z k on both sides of this equation gives:
Convolution yields
1
1
1 9
1
8
1
7
1
6
28
1
5
21
36 84
1
4
15
56
1
3
10
35
1
2
6
20
70
Vandermonde’s convolution:
k
m n m + n
∑
=
j=0 j k − j k
1
3
10
35
126 126
1
4
15
56
1
5
21
1
6
28
1
7
84 36
1
8
1
1
9 1
9 3 6 3 6 3 6
= + +
2 0 2 1 1 2 0

Vandermonde’s identity (3)
Special case m = n = k yields:
2n
=
n
n
∑
j=0
Example: 42 42 42 42 42 8 0 + 1 + 2 + 3 + 4 = 4
1
1
1 8
1
7
1
6
1
5
21
1
4
15
28 56
1
3
10
35
1
2
6
20
1
3
10
35
1
4
15
70 56
1
5
21
1
6
2 n n n
=
j n − j ∑
j0 j
1
7
1
28 8
1
1
1
1
1 16
1
9
1
4
1
9
1
36 16
1 + 16 + 36 + 16 + 1 = 70
1
1

Vandermonde’s identity (3)
Special case m = n = k yields:
2n
=
n
n
∑
j=0
Example: 42 42 42 42 42 8 0 + 1 + 2 + 3 + 4 = 4
1
1
1 8
1
7
1
6
1
5
21
1
4
15
28 56
1
3
10
35
1
2
6
20
1
3
10
35
1
4
15
70 56
1
5
21
1
6
2 n n n
=
j n − j ∑
j0 j
1
7
1
28 8
1
1
1
1
1 16
1
9
1
4
1
9
1
36 16
1 + 16 + 36 + 16 + 1 = 70
1
1

Sequence m
0
Let’s take sequences
and
,0,− m
1
,0, m
2
,0,− m
3
,0, m
4
m m m m
〈 , , ,..., ,...〉 ←→ F (X ) = (1 + x)
0 1 2 n
m
,0,...,
m m m m
〈 ,− , ,− ,...,(−1)
0 1 2 3
n
m
,...〉 ←→ G(X ) = (1 − x)
n
m
Then the convolution corresponds to the function (1 − x) m (1 + x) m = (1 − x 2 ) m that
gives the identity of binomial coefficients:
n
m m
∑
(−1)
j=0 j n − j
j = (−1) n/2
m
[n is even].
n/2

Some other useful sequences
For any n 0:
1
(1 − x) n+1 = ∑
k0
−n − 1
k
(−x) k = ∑
k0
= (−1) k n−k−1 k
Here the identity negative coefficients n k
For any n 0, using right-shift :
xn (1 − x) n+1 =
k
∑ x
k0 n
k
Exponential series
Logarithmic series
n + k
e x = 1 + x + 1
2! x2 + 1
3! x3 1
+ ··· = ∑
k0 k! xk
ln(1 + x) = x − 1
2 x2 + 1
3 x3 − ··· = ∑
k1
n
x k
is used.
(−1) k+1
x
k
k
1 1
= x +
ln(1 + x) 2 x2 + 1
3 x3 1
+ ··· = ∑
k1 k xk