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Combinatorics Pascal’s 4 Proofs ???
Combinatorial Enumeration in Pascal’s TriangleHow To Count Without Counting
Brian K. Miceli
Trinity UniversityMathematics Department
Mathematics Majors’ SeminarMarch 23, 2016
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Outline
1 What is Combinatorics?
2 Pascal’s Triangle
3 Proofs
4 One Last Thing
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Outline
1 What is Combinatorics?
2 Pascal’s Triangle
3 Proofs
4 One Last Thing
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Outline
1 What is Combinatorics?
2 Pascal’s Triangle
3 Proofs
4 One Last Thing
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Outline
1 What is Combinatorics?
2 Pascal’s Triangle
3 Proofs
4 One Last Thing
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
What is Combinatorics?
Combinatorics (MATH 3343) is a class offered in the fall of 2016.
As a branch of mathematics, I like to say that it’s the Art ofCounting, and it has its own style of proof, called (unsurprisingly)the combinatorial proof.
This is best illustrated with an example.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
What is Combinatorics?
Combinatorics (MATH 3343) is a class offered in the fall of 2016.
As a branch of mathematics, I like to say that it’s the Art ofCounting, and it has its own style of proof, called (unsurprisingly)the combinatorial proof.
This is best illustrated with an example.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
What is Combinatorics?
Combinatorics (MATH 3343) is a class offered in the fall of 2016.
As a branch of mathematics, I like to say that it’s the Art ofCounting, and it has its own style of proof, called (unsurprisingly)the combinatorial proof.
This is best illustrated with an example.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
What is Combinatorics?
Combinatorics (MATH 3343) is a class offered in the fall of 2016.
As a branch of mathematics, I like to say that it’s the Art ofCounting, and it has its own style of proof, called (unsurprisingly)the combinatorial proof.
This is best illustrated with an example.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
A Typical Combinatorial Problem
Theorem
For any n ∈ N, n + (n − 1) + (n − 2) + · · ·+ 2 + 1 =(n + 1)n
2.
Proof: (A combinatorial proof.) Let’s imagine that we have agroup of n + 1 people–Albert (A), Brian (B), . . ., Norbert (N),Oscar (O=N+1)–who enter a contest with two identical grandprizes. How many ways can we choose the two people who win?(i) We can choose any of the n + 1 people to win in the firstdrawing, and then choose one of the remaining n people to win inthe second. There are (n + 1)n ways to do this. But thisdifferentiates between choosing A and then B and choosing B andthen A, which is irrelevant, so we must divide by 2. Thus, the
number of ways of choosing the winners is(n + 1)n
2.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
A Typical Combinatorial Problem
Theorem
For any n ∈ N, n + (n − 1) + (n − 2) + · · ·+ 2 + 1 =(n + 1)n
2.
Proof: (A combinatorial proof.) Let’s imagine that we have agroup of n + 1 people–Albert (A), Brian (B), . . ., Norbert (N),Oscar (O=N+1)–who enter a contest with two identical grandprizes. How many ways can we choose the two people who win?(i) We can choose any of the n + 1 people to win in the firstdrawing, and then choose one of the remaining n people to win inthe second. There are (n + 1)n ways to do this. But thisdifferentiates between choosing A and then B and choosing B andthen A, which is irrelevant, so we must divide by 2. Thus, the
number of ways of choosing the winners is(n + 1)n
2.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
A Typical Combinatorial Problem
Theorem
For any n ∈ N, n + (n − 1) + (n − 2) + · · ·+ 2 + 1 =(n + 1)n
2.
Proof: (A combinatorial proof.) Let’s imagine that we have agroup of n + 1 people–Albert (A), Brian (B), . . ., Norbert (N),Oscar (O=N+1)–who enter a contest with two identical grandprizes. How many ways can we choose the two people who win?
(i) We can choose any of the n + 1 people to win in the firstdrawing, and then choose one of the remaining n people to win inthe second. There are (n + 1)n ways to do this. But thisdifferentiates between choosing A and then B and choosing B andthen A, which is irrelevant, so we must divide by 2. Thus, the
number of ways of choosing the winners is(n + 1)n
2.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
A Typical Combinatorial Problem
Theorem
For any n ∈ N, n + (n − 1) + (n − 2) + · · ·+ 2 + 1 =(n + 1)n
2.
Proof: (A combinatorial proof.) Let’s imagine that we have agroup of n + 1 people–Albert (A), Brian (B), . . ., Norbert (N),Oscar (O=N+1)–who enter a contest with two identical grandprizes. How many ways can we choose the two people who win?(i) We can choose any of the n + 1 people to win in the firstdrawing, and then choose one of the remaining n people to win inthe second. There are (n + 1)n ways to do this.
But thisdifferentiates between choosing A and then B and choosing B andthen A, which is irrelevant, so we must divide by 2. Thus, the
number of ways of choosing the winners is(n + 1)n
2.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
A Typical Combinatorial Problem
Theorem
For any n ∈ N, n + (n − 1) + (n − 2) + · · ·+ 2 + 1 =(n + 1)n
2.
Proof: (A combinatorial proof.) Let’s imagine that we have agroup of n + 1 people–Albert (A), Brian (B), . . ., Norbert (N),Oscar (O=N+1)–who enter a contest with two identical grandprizes. How many ways can we choose the two people who win?(i) We can choose any of the n + 1 people to win in the firstdrawing, and then choose one of the remaining n people to win inthe second. There are (n + 1)n ways to do this. But thisdifferentiates between choosing A and then B and choosing B andthen A, which is irrelevant, so we must divide by 2.
Thus, the
number of ways of choosing the winners is(n + 1)n
2.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
A Typical Combinatorial Problem
Theorem
For any n ∈ N, n + (n − 1) + (n − 2) + · · ·+ 2 + 1 =(n + 1)n
2.
Proof: (A combinatorial proof.) Let’s imagine that we have agroup of n + 1 people–Albert (A), Brian (B), . . ., Norbert (N),Oscar (O=N+1)–who enter a contest with two identical grandprizes. How many ways can we choose the two people who win?(i) We can choose any of the n + 1 people to win in the firstdrawing, and then choose one of the remaining n people to win inthe second. There are (n + 1)n ways to do this. But thisdifferentiates between choosing A and then B and choosing B andthen A, which is irrelevant, so we must divide by 2. Thus, the
number of ways of choosing the winners is(n + 1)n
2.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
A Typical Combinatorial ProblemThe proof continued . . .
(ii) Now organize the winning pairs by who is alphabetically first.That is, the number of winning pairs is
#(A is 1st) + #(B is 1st) + · · ·+ #(N is 1st) + #(O is 1st),
which isn + (n − 1) + · · ·+ 1 + 0,
which completes the proof.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
A Typical Combinatorial ProblemThe proof continued . . .
(ii) Now organize the winning pairs by who is alphabetically first.
That is, the number of winning pairs is
#(A is 1st) + #(B is 1st) + · · ·+ #(N is 1st) + #(O is 1st),
which isn + (n − 1) + · · ·+ 1 + 0,
which completes the proof.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
A Typical Combinatorial ProblemThe proof continued . . .
(ii) Now organize the winning pairs by who is alphabetically first.That is, the number of winning pairs is
#(A is 1st) + #(B is 1st) + · · ·+ #(N is 1st) + #(O is 1st),
which isn + (n − 1) + · · ·+ 1 + 0,
which completes the proof.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
A Typical Combinatorial ProblemThe proof continued . . .
(ii) Now organize the winning pairs by who is alphabetically first.That is, the number of winning pairs is
#(A is 1st) + #(B is 1st) + · · ·+ #(N is 1st) + #(O is 1st),
which isn + (n − 1) + · · ·+ 1 + 0,
which completes the proof.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Some Basic Combinatorial ObjectsDefinitions
We define
(n
k
)=
n!
k!(n − k)!=
n(n − 1) · · · (n − k + 1)
k!,
where n! = n(n − 1) · · · (2)(1).
Then n! corresponds to the number of ways of ordering n distinctobjects.
We then have that(nk
)corresponds to the number of k-element
subsets of an n-element set. By this definition, we also let(nk
)= 0
if n < 0, k < 0, or k > n.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Some Basic Combinatorial ObjectsDefinitions
We define
(n
k
)=
n!
k!(n − k)!=
n(n − 1) · · · (n − k + 1)
k!,
where n! = n(n − 1) · · · (2)(1).
Then n! corresponds to the number of ways of ordering n distinctobjects.
We then have that(nk
)corresponds to the number of k-element
subsets of an n-element set. By this definition, we also let(nk
)= 0
if n < 0, k < 0, or k > n.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Some Basic Combinatorial ObjectsDefinitions
We define
(n
k
)=
n!
k!(n − k)!=
n(n − 1) · · · (n − k + 1)
k!,
where n! = n(n − 1) · · · (2)(1).
Then n! corresponds to the number of ways of ordering n distinctobjects.
We then have that(nk
)corresponds to the number of k-element
subsets of an n-element set. By this definition, we also let(nk
)= 0
if n < 0, k < 0, or k > n.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Some Basic Combinatorial ObjectsDefinitions
We define
(n
k
)=
n!
k!(n − k)!=
n(n − 1) · · · (n − k + 1)
k!,
where n! = n(n − 1) · · · (2)(1).
Then n! corresponds to the number of ways of ordering n distinctobjects.
We then have that(nk
)corresponds to the number of k-element
subsets of an n-element set.
By this definition, we also let(nk
)= 0
if n < 0, k < 0, or k > n.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Some Basic Combinatorial ObjectsDefinitions
We define
(n
k
)=
n!
k!(n − k)!=
n(n − 1) · · · (n − k + 1)
k!,
where n! = n(n − 1) · · · (2)(1).
Then n! corresponds to the number of ways of ordering n distinctobjects.
We then have that(nk
)corresponds to the number of k-element
subsets of an n-element set. By this definition, we also let(nk
)= 0
if n < 0, k < 0, or k > n.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Some Basic Combinatorial ObjectsExamples
3! = 3 · 2 · 1 = 6, so there should be six ways of ordering theelements of the set {a, b, c}:
abc, acb, bac, bca, cab, cba.
(4
2
)=
4!
2!(4− 2)!= 6, so there should be six 2-element subsets of
the set {A,B,C ,D}:
{A,B}, {A,C}, {A,D}, {B,C}, {B,D}, {C ,D}.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Some Basic Combinatorial ObjectsExamples
3! = 3 · 2 · 1 = 6, so there should be six ways of ordering theelements of the set {a, b, c}:
abc, acb, bac, bca, cab, cba.
(4
2
)=
4!
2!(4− 2)!= 6, so there should be six 2-element subsets of
the set {A,B,C ,D}:
{A,B}, {A,C}, {A,D}, {B,C}, {B,D}, {C ,D}.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Some Basic Combinatorial ObjectsExamples
3! = 3 · 2 · 1 = 6, so there should be six ways of ordering theelements of the set {a, b, c}:
abc, acb, bac, bca, cab, cba.
(4
2
)=
4!
2!(4− 2)!= 6, so there should be six 2-element subsets of
the set {A,B,C ,D}:
{A,B}, {A,C}, {A,D}, {B,C}, {B,D}, {C ,D}.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Some Basic Combinatorial ObjectsExamples
3! = 3 · 2 · 1 = 6, so there should be six ways of ordering theelements of the set {a, b, c}:
abc, acb, bac, bca, cab, cba.
(4
2
)=
4!
2!(4− 2)!= 6, so there should be six 2-element subsets of
the set {A,B,C ,D}:
{A,B}, {A,C}, {A,D}, {B,C}, {B,D}, {C ,D}.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Some Basic Combinatorial ObjectsExamples
3! = 3 · 2 · 1 = 6, so there should be six ways of ordering theelements of the set {a, b, c}:
abc, acb, bac, bca, cab, cba.
(4
2
)=
4!
2!(4− 2)!= 6, so there should be six 2-element subsets of
the set {A,B,C ,D}:
{A,B}, {A,C}, {A,D}, {B,C}, {B,D}, {C ,D}.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
A Key Combinatorial IdentityTwo Proofs
Theorem
For any n, k ∈ N,(n
k
)=
(n − 1
k
)+
(n − 1
k − 1
).
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
A Key Combinatorial IdentityTwo Proofs
Theorem
For any n, k ∈ N,(n
k
)=
(n − 1
k
)+
(n − 1
k − 1
).
Proof: (An algebraic proof.)(n − 1
k
)+
(n − 1
k − 1
)=
(n − 1)!
k!(n − 1− k)!+
(n − 1)!
(k − 1)!(n − k)!
=(n − k)(n − 1)!
k!(n − 1− k)!(n − k)+
k(n − 1)!
k(k − 1)!(n − k)!
=(n − k)(n − 1)! + k(n − 1)!
k!(n − k)!
=n(n − 1)!
k!(n − k)!=
(n
k
)
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
A Key Combinatorial IdentityTwo Proofs
Theorem
For any n, k ∈ N,(n
k
)=
(n − 1
k
)+
(n − 1
k − 1
).
Proof: (A combinatorial proof.) How many ways are there tochoose a k-element subset from the set {1, 2, . . . n}?(i) By definition, there are
(nk
)ways to do this.
(ii) Alternatively, we can divide all such subsets into two cases bywhether or not our k-element subsets contain n. If n is not in oursubset, then we must choose our k elements from the set{1, 2, . . . , n − 1}, and there are
(n−1k
)ways to do that. If n is in
our subset, then we must choose the other k − 1 elements of oursubset from the set {1, 2, . . . , n − 1}, and there are
(n−1k−1
)ways to
do that.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
A Key Combinatorial IdentityTwo Proofs
Theorem
For any n, k ∈ N,(n
k
)=
(n − 1
k
)+
(n − 1
k − 1
).
Proof: (A combinatorial proof.) How many ways are there tochoose a k-element subset from the set {1, 2, . . . n}?(i) By definition, there are
(nk
)ways to do this.
(ii) Alternatively, we can divide all such subsets into two cases bywhether or not our k-element subsets contain n. f n is not in oursubset, then we must choose our k elements from the set{1, 2, . . . , n − 1}, and there are
(n−1k
)ways to do that. If n is in
our subset, then we must choose the other k − 1 elements of oursubset from the set {1, 2, . . . , n − 1}, and there are
(n−1k−1
)ways to
do that.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
A Key Combinatorial IdentityTwo Proofs
Theorem
For any n, k ∈ N,(n
k
)=
(n − 1
k
)+
(n − 1
k − 1
).
Proof: (A combinatorial proof.) How many ways are there tochoose a k-element subset from the set {1, 2, . . . n}?(i) By definition, there are
(nk
)ways to do this.
(ii) Alternatively, we can divide all such subsets into two cases bywhether or not our k-element subsets contain n. If n is not in oursubset, then we must choose our k elements from the set{1, 2, . . . , n − 1}, and there are
(n−1k
)ways to do that. If n is in
our subset, then we must choose the other k − 1 elements of oursubset from the set {1, 2, . . . , n − 1}, and there are
(n−1k−1
)ways to
do that.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
A Key Combinatorial IdentityTwo Proofs
Theorem
For any n, k ∈ N,(n
k
)=
(n − 1
k
)+
(n − 1
k − 1
).
Proof: (A combinatorial proof.) How many ways are there tochoose a k-element subset from the set {1, 2, . . . n}?(i) By definition, there are
(nk
)ways to do this.
(ii) Alternatively, we can divide all such subsets into two cases bywhether or not our k-element subsets contain n. If n is not in oursubset, then we must choose our k elements from the set{1, 2, . . . , n − 1}, and there are
(n−1k
)ways to do that. If n is in
our subset, then we must choose the other k − 1 elements of oursubset from the set {1, 2, . . . , n − 1}, and there are
(n−1k−1
)ways to
do that.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
A Key Combinatorial IdentityTwo Proofs
Theorem
For any n, k ∈ N,(n
k
)=
(n − 1
k
)+
(n − 1
k − 1
).
Proof: (A combinatorial proof.) How many ways are there tochoose a k-element subset from the set {1, 2, . . . n}?(i) By definition, there are
(nk
)ways to do this.
(ii) Alternatively, we can divide all such subsets into two cases bywhether or not our k-element subsets contain n. If n is not in oursubset, then we must choose our k elements from the set{1, 2, . . . , n − 1}, and there are
(n−1k
)ways to do that. If n is in
our subset, then we must choose the other k − 1 elements of oursubset from the set {1, 2, . . . , n − 1}, and there are
(n−1k−1
)ways to
do that.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4
Pascal’s Triangle is named after Blaise Pascal, a Frenchmathematician in the 1600’s.
Pascal’s Triangle was known in China as early as the late 1200’s.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4
Pascal’s Triangle is named after Blaise Pascal, a Frenchmathematician in the 1600’s.
Pascal’s Triangle was known in China as early as the late 1200’s.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4
Pascal’s Triangle is named after Blaise Pascal, a Frenchmathematician in the 1600’s.
Pascal’s Triangle was known in China as early as the late 1200’s.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4
Pascal’s Triangle is named after Blaise Pascal, a Frenchmathematician in the 1600’s.
Pascal’s Triangle was known in China as early as the late 1200’s.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 411 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 411 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
Each entry is the sum of two numbers: the one directly above andthe one to the left of that one, where we assume any blank spaceis a 0.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 411 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
We refer to the element in the nth row and kth column as Pn,k .
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 411 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
We refer to the element in the nth row and kth column as Pn,k .For example, P3,2 = 3,
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 411 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
We refer to the element in the nth row and kth column as Pn,k .For example, P3,2 = 3, P4,0 = 1,
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 411 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
We refer to the element in the nth row and kth column as Pn,k .For example, P3,2 = 3, P4,0 = 1, P5,3 = 10,
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 411 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
We refer to the element in the nth row and kth column as Pn,k .For example, P3,2 = 3, P4,0 = 1, P5,3 = 10, P6,9 = 0.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 411 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
By the known relation in Pascal’s Triangle,Pn,k = Pn−1,k + Pn−1,k−1 for any n, k ∈ Z.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4
Theorem
For every n, k ∈ Z,(n
k
)= Pn,k .
Proof: As stated, Pn,k = Pn−1,k + Pn−1,k−1, with Pn,k = 0 ifn < 0, k < 0, or n < k , and from Pascal’s Triangle, P0,0 = 1.
We can compute that(00
)= 1, and we have that
(nk
)= 0 whenever
n < 0, k < 0, or n < k .
Since(nk
)=(n−1
k
)+(n−1k−1
), we see that Pn,k and
(nk
)satisfy the
same recursions and initial conditions, so it must be the case that(nk
)= Pn,k for every n, k ∈ Z.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4
Theorem
For every n, k ∈ Z,(n
k
)= Pn,k .
Proof: As stated, Pn,k = Pn−1,k + Pn−1,k−1, with Pn,k = 0 ifn < 0, k < 0, or n < k , and from Pascal’s Triangle, P0,0 = 1.
We can compute that(00
)= 1, and we have that
(nk
)= 0 whenever
n < 0, k < 0, or n < k .
Since(nk
)=(n−1
k
)+(n−1k−1
), we see that Pn,k and
(nk
)satisfy the
same recursions and initial conditions, so it must be the case that(nk
)= Pn,k for every n, k ∈ Z.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4
Theorem
For every n, k ∈ Z,(n
k
)= Pn,k .
Proof: As stated, Pn,k = Pn−1,k + Pn−1,k−1, with Pn,k = 0 ifn < 0, k < 0, or n < k , and from Pascal’s Triangle, P0,0 = 1.
We can compute that(00
)= 1, and we have that
(nk
)= 0 whenever
n < 0, k < 0, or n < k .
Since(nk
)=(n−1
k
)+(n−1k−1
), we see that Pn,k and
(nk
)satisfy the
same recursions and initial conditions, so it must be the case that(nk
)= Pn,k for every n, k ∈ Z.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4
Theorem
For every n, k ∈ Z,(n
k
)= Pn,k .
Proof: As stated, Pn,k = Pn−1,k + Pn−1,k−1, with Pn,k = 0 ifn < 0, k < 0, or n < k , and from Pascal’s Triangle, P0,0 = 1.
We can compute that(00
)= 1, and we have that
(nk
)= 0 whenever
n < 0, k < 0, or n < k .
Since(nk
)=(n−1
k
)+(n−1k−1
), we see that Pn,k and
(nk
)satisfy the
same recursions and initial conditions, so it must be the case that(nk
)= Pn,k for every n, k ∈ Z.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4
Theorem
For every n, k ∈ Z,(n
k
)= Pn,k .
Proof: As stated, Pn,k = Pn−1,k + Pn−1,k−1, with Pn,k = 0 ifn < 0, k < 0, or n < k , and from Pascal’s Triangle, P0,0 = 1.
We can compute that(00
)= 1, and we have that
(nk
)= 0 whenever
n < 0, k < 0, or n < k .
Since(nk
)=(n−1
k
)+(n−1k−1
), we see that Pn,k and
(nk
)satisfy the
same recursions and initial conditions, so it must be the case that(nk
)= Pn,k for every n, k ∈ Z.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4
In other words, Pascal’s Triangle looks like this:(00
)(10
) (11
)(20
) (21
) (22
)(30
) (31
) (32
) (33
)(40
) (41
) (42
) (43
) (44
)(50
) (51
) (52
) (53
) (54
) (55
)(60
) (61
) (62
) (63
) (64
) (65
) (66
)Keeping this in mind, let’s see what we can conjecture about thenumbers in Pascal’s Triangle . . .
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4
In other words, Pascal’s Triangle looks like this:
(00
)(10
) (11
)(20
) (21
) (22
)(30
) (31
) (32
) (33
)(40
) (41
) (42
) (43
) (44
)(50
) (51
) (52
) (53
) (54
) (55
)(60
) (61
) (62
) (63
) (64
) (65
) (66
)Keeping this in mind, let’s see what we can conjecture about thenumbers in Pascal’s Triangle . . .
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4
In other words, Pascal’s Triangle looks like this:(00
)(10
) (11
)(20
) (21
) (22
)(30
) (31
) (32
) (33
)(40
) (41
) (42
) (43
) (44
)(50
) (51
) (52
) (53
) (54
) (55
)(60
) (61
) (62
) (63
) (64
) (65
) (66
)
Keeping this in mind, let’s see what we can conjecture about thenumbers in Pascal’s Triangle . . .
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4
In other words, Pascal’s Triangle looks like this:(00
)(10
) (11
)(20
) (21
) (22
)(30
) (31
) (32
) (33
)(40
) (41
) (42
) (43
) (44
)(50
) (51
) (52
) (53
) (54
) (55
)(60
) (61
) (62
) (63
) (64
) (65
) (66
)Keeping this in mind, let’s see what we can conjecture about thenumbers in Pascal’s Triangle . . .
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4Interesting Properties
11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4Interesting Properties
11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
What do we get when sum across the rows of Pascal’s Triangle?
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4Interesting Properties
11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
What do we get if we sum from the tops of the columns and godown to some stopping point?
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4Interesting Properties
11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
What do we get if we take an alternating sum across the rows?
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4Interesting Properties
11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
What do we get if we sum diagonally, starting at the far right andworking down and left?
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4Interesting Properties
11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4Interesting Properties
11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
What do we get when sum across the rows of Pascal’s Triangle?
n∑k=0
(n
k
)= 2n
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4Interesting Properties
11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
What do we get when sum across the rows of Pascal’s Triangle?
n∑k=0
(n
k
)= 2n
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4Interesting Properties
11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
What do we get if we sum from the tops of the columns and godown to some stopping point?
n∑i=k
(i
k
)=
(n + 1
k + 1
)
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4Interesting Properties
11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
What do we get if we sum from the tops of the columns and godown to some stopping point?
n∑i=k
(i
k
)=
(n + 1
k + 1
)
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4Interesting Properties
11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
What do we get if we take an alternating sum across the rows?
n∑k=0
(−1)k(n
k
)= 0
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4Interesting Properties
11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
What do we get if we take an alternating sum across the rows?
n∑k=0
(−1)k(n
k
)= 0
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4Interesting Properties
11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
What do we get if we sum diagonally, starting at the far right andworking down and left? ∑
i+j=n,i≤j
(j
i
)= fn
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Pascal’s 4Interesting Properties
11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1
What do we get if we sum diagonally, starting at the far right andworking down and left? ∑
i+j=n,i≤j
(j
i
)= fn
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
A Proof of the First Conjecture
Theorem
For n ≥ 0,n∑
k=0
(n
k
)= 2n.
Proof: How many subsets of A = {1, 2, . . . , n} exist?
(i) From Intro to Abstract, we know there are 2n subsets of A.
(ii) The total number of subsets is also equal to
#(0-elmt. subsets)+#(1-elmt. subsets)+· · ·+#(n-elmt. subsets),
which is (n
0
)+
(n
1
)+ · · ·+
(n
n
).
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
A Proof of the First Conjecture
Theorem
For n ≥ 0,n∑
k=0
(n
k
)= 2n.
Proof: How many subsets of A = {1, 2, . . . , n} exist?
(i) From Intro to Abstract, we know there are 2n subsets of A.
(ii) The total number of subsets is also equal to
#(0-elmt. subsets)+#(1-elmt. subsets)+· · ·+#(n-elmt. subsets),
which is (n
0
)+
(n
1
)+ · · ·+
(n
n
).
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
A Proof of the First Conjecture
Theorem
For n ≥ 0,n∑
k=0
(n
k
)= 2n.
Proof: How many subsets of A = {1, 2, . . . , n} exist?
(i) From Intro to Abstract, we know there are 2n subsets of A.
(ii) The total number of subsets is also equal to
#(0-elmt. subsets)+#(1-elmt. subsets)+· · ·+#(n-elmt. subsets),
which is (n
0
)+
(n
1
)+ · · ·+
(n
n
).
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
A Proof of the First Conjecture
Theorem
For n ≥ 0,n∑
k=0
(n
k
)= 2n.
Proof: How many subsets of A = {1, 2, . . . , n} exist?
(i) From Intro to Abstract, we know there are 2n subsets of A.
(ii) The total number of subsets is also equal to
#(0-elmt. subsets)+#(1-elmt. subsets)+· · ·+#(n-elmt. subsets),
which is (n
0
)+
(n
1
)+ · · ·+
(n
n
).
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
A Proof of the Third Conjecture
Theorem
For n ≥ 1,n∑
k=0
(−1)k(n
k
)= 0.
Before we begin the proof, let’s think about what this theorem isasking us to show.
In terms of numbers of subsets of A = {1, 2, . . . , n}, we want toshow
#(even subsets of A)−#(odd subsets of A) = 0,
or#(even subsets of A) = #(odd subsets of A).
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
A Proof of the Third Conjecture
Theorem
For n ≥ 1,n∑
k=0
(−1)k(n
k
)= 0.
Before we begin the proof, let’s think about what this theorem isasking us to show.
In terms of numbers of subsets of A = {1, 2, . . . , n}, we want toshow
#(even subsets of A)−#(odd subsets of A) = 0,
or#(even subsets of A) = #(odd subsets of A).
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
A Proof of the Third Conjecture
Theorem
For n ≥ 1,n∑
k=0
(−1)k(n
k
)= 0.
Before we begin the proof, let’s think about what this theorem isasking us to show.
In terms of numbers of subsets of A = {1, 2, . . . , n}, we want toshow
#(even subsets of A)−#(odd subsets of A) = 0,
or#(even subsets of A) = #(odd subsets of A).
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
A Proof of the Third Conjecture
Theorem
For n ≥ 1,n∑
k=0
(−1)k(n
k
)= 0.
Before we begin the proof, let’s think about what this theorem isasking us to show.
In terms of numbers of subsets of A = {1, 2, . . . , n}, we want toshow
#(even subsets of A)−#(odd subsets of A) = 0,
or#(even subsets of A) = #(odd subsets of A).
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
A Proof of the Third Conjecture
Theorem
For n ≥ 1,n∑
k=0
(−1)k(n
k
)= 0.
Before we begin the proof, let’s think about what this theorem isasking us to show.
In terms of numbers of subsets of A = {1, 2, . . . , n}, we want toshow
#(even subsets of A)−#(odd subsets of A) = 0,
or#(even subsets of A) = #(odd subsets of A).
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
A Proof of the Third Conjecture
Proof: Let En denote the even subsets of A, and let On denote theodd subset of A.
Define f : En → On by
i. f (X ) = X ∪ {1} if 1 /∈ X , and
ii. f (X ) = X − {1} if 1 ∈ X .
Then f is invertible, so |En| = |On|.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
A Proof of the Third Conjecture
Proof: Let En denote the even subsets of A, and let On denote theodd subset of A.
Define f : En → On by
i. f (X ) = X ∪ {1} if 1 /∈ X , and
ii. f (X ) = X − {1} if 1 ∈ X .
Then f is invertible, so |En| = |On|.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
A Proof of the Third Conjecture
Proof: Let En denote the even subsets of A, and let On denote theodd subset of A.
Define f : En → On by
i. f (X ) = X ∪ {1} if 1 /∈ X , and
ii. f (X ) = X − {1} if 1 ∈ X .
Then f is invertible, so |En| = |On|.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
A Proof of the Third Conjecture
Proof: Let En denote the even subsets of A, and let On denote theodd subset of A.
Define f : En → On by
i. f (X ) = X ∪ {1} if 1 /∈ X , and
ii. f (X ) = X − {1} if 1 ∈ X .
Then f is invertible, so |En| = |On|.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Proofs of the Other Conjectures
Theorem
For n ≥ 0,n∑
i=k
(i
k
)=
(n + 1
k + 1
).
Theorem
For n ≥ 0,∑
i+j=n,i≤j
(j
i
)= fn.
Proof: Take 3343 in the fall.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Proofs of the Other Conjectures
Theorem
For n ≥ 0,n∑
i=k
(i
k
)=
(n + 1
k + 1
).
Theorem
For n ≥ 0,∑
i+j=n,i≤j
(j
i
)= fn.
Proof: Take 3343 in the fall.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Proofs of the Other Conjectures
Theorem
For n ≥ 0,n∑
i=k
(i
k
)=
(n + 1
k + 1
).
Theorem
For n ≥ 0,∑
i+j=n,i≤j
(j
i
)= fn.
Proof:
Take 3343 in the fall.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Proofs of the Other Conjectures
Theorem
For n ≥ 0,n∑
i=k
(i
k
)=
(n + 1
k + 1
).
Theorem
For n ≥ 0,∑
i+j=n,i≤j
(j
i
)= fn.
Proof: Take 3343 in the fall.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Something to Recall from 1312
Let g(x) =∑n≥0
xn = 1 + x + x2 + x3 + · · · .
In Calc II we learn that g(x) =1
1− x, provided that |x | < 1.
One way to think of this is that g(x) is the Maclaurin series for1
1−x , which has an interval of convergence of (−1, 1).
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Something to Recall from 1312
Let g(x) =∑n≥0
xn = 1 + x + x2 + x3 + · · · .
In Calc II we learn that g(x) =1
1− x, provided that |x | < 1.
One way to think of this is that g(x) is the Maclaurin series for1
1−x , which has an interval of convergence of (−1, 1).
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Something to Recall from 1312
Let g(x) =∑n≥0
xn = 1 + x + x2 + x3 + · · · .
In Calc II we learn that g(x) =1
1− x, provided that |x | < 1.
One way to think of this is that g(x) is the Maclaurin series for1
1−x , which has an interval of convergence of (−1, 1).
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Something to Recall from 1312
Let g(x) =∑n≥0
xn = 1 + x + x2 + x3 + · · · .
In Calc II we learn that g(x) =1
1− x, provided that |x | < 1.
One way to think of this is that g(x) is the Maclaurin series for1
1−x , which has an interval of convergence of (−1, 1).
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Why Does This Happen?
What is the Maclaurin series for F (x) =1
1− x − x2?
F (x) = 1 + x + 2x2 + 3x3 + 5x4 + 8x5 + 13x6 + 21x7 + 34x8 + · · ·=
∑n≥0
fnxn.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Why Does This Happen?
What is the Maclaurin series for F (x) =1
1− x − x2?
F (x) = 1 + x + 2x2 + 3x3 + 5x4 + 8x5 + 13x6 + 21x7 + 34x8 + · · ·
=∑n≥0
fnxn.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
Why Does This Happen?
What is the Maclaurin series for F (x) =1
1− x − x2?
F (x) = 1 + x + 2x2 + 3x3 + 5x4 + 8x5 + 13x6 + 21x7 + 34x8 + · · ·=
∑n≥0
fnxn.
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
The End
Thanks for listening.
Questions?
Miceli Combinatorics
Combinatorics Pascal’s 4 Proofs ???
The End
Thanks for listening.Questions?
Miceli Combinatorics
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