Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count

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


Outline

1 What is Combinatorics?

2 Pascal’s Triangle

3 Proofs

4 One Last Thing



Outline



3 Proofs

4 One Last Thing



Outline



3 Proofs

4 One Last Thing



Outline



3 Proofs

4 One Last Thing



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.





















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.




Theorem

For any n ∈ N, n + (n − 1) + (n − 2) + · · ·+ 2 + 1 =(n + 1)n

2.



2.




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


2.




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


2.




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


2.




Theorem

For any n ∈ N, n + (n − 1) + (n − 2) + · · ·+ 2 + 1 =(n + 1)n

2.



2.



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.




(ii) Now organize the winning pairs by who is alphabetically first.

That is, the number of winning pairs is


which isn + (n − 1) + · · ·+ 1 + 0,







which isn + (n − 1) + · · ·+ 1 + 0,







which isn + (n − 1) + · · ·+ 1 + 0,




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.




We define

(n

k

)=

n!

k!(n − k)!=

n(n − 1) · · · (n − k + 1)

k!,

where n! = n(n − 1) · · · (2)(1).





)= 0

if n < 0, k < 0, or k > n.




We define

(n

k

)=

n!

k!(n − k)!=

n(n − 1) · · · (n − k + 1)

k!,

where n! = n(n − 1) · · · (2)(1).





)= 0

if n < 0, k < 0, or k > n.




We define

(n

k

)=

n!

k!(n − k)!=

n(n − 1) · · · (n − k + 1)

k!,

where n! = n(n − 1) · · · (2)(1).




subsets of an n-element set.

By this definition, we also let(nk

)= 0

if n < 0, k < 0, or k > n.




We define

(n

k

)=

n!

k!(n − k)!=

n(n − 1) · · · (n − k + 1)

k!,

where n! = n(n − 1) · · · (2)(1).





)= 0

if n < 0, k < 0, or k > n.



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}.






(4

2

)=

4!


the set {A,B,C ,D}:

{A,B}, {A,C}, {A,D}, {B,C}, {B,D}, {C ,D}.






(4

2

)=

4!


the set {A,B,C ,D}:

{A,B}, {A,C}, {A,D}, {B,C}, {B,D}, {C ,D}.






(4

2

)=

4!


the set {A,B,C ,D}:

{A,B}, {A,C}, {A,D}, {B,C}, {B,D}, {C ,D}.






(4

2

)=

4!


the set {A,B,C ,D}:

{A,B}, {A,C}, {A,D}, {B,C}, {B,D}, {C ,D}.



A Key Combinatorial IdentityTwo Proofs

Theorem

For any n, k ∈ N,(n

k

)=

(n − 1

k

)+

(n − 1

k − 1

).




Theorem


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

)




Theorem


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.




Theorem


k

)=

(n − 1

k

)+

(n − 1

k − 1

).


(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



(n−1k−1

)ways to

do that.




Theorem


k

)=

(n − 1

k

)+

(n − 1

k − 1

).


(nk

)ways to do this.


(n−1k



(n−1k−1

)ways to

do that.




Theorem


k

)=

(n − 1

k

)+

(n − 1

k − 1

).


(nk

)ways to do this.


(n−1k



(n−1k−1

)ways to

do that.




Theorem


k

)=

(n − 1

k

)+

(n − 1

k − 1

).


(nk

)ways to do this.


(n−1k



(n−1k−1

)ways to

do that.



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.



Pascal’s 4





Pascal’s 4





Pascal’s 4





Pascal’s 4



Pascal’s 411 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1



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.



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 .



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,



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,



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,



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.



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.



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.



Pascal’s 4

Theorem


k

)= Pn,k .




(nk

)= 0 whenever

n < 0, k < 0, or n < k .

Since(nk

)=(n−1

k

)+(n−1k−1


(nk

)satisfy the





Pascal’s 4

Theorem


k

)= Pn,k .




(nk

)= 0 whenever

n < 0, k < 0, or n < k .

Since(nk

)=(n−1

k

)+(n−1k−1


(nk

)satisfy the





Pascal’s 4

Theorem


k

)= Pn,k .




(nk

)= 0 whenever

n < 0, k < 0, or n < k .

Since(nk

)=(n−1

k

)+(n−1k−1


(nk

)satisfy the





Pascal’s 4

Theorem


k

)= Pn,k .




(nk

)= 0 whenever

n < 0, k < 0, or n < k .

Since(nk

)=(n−1

k

)+(n−1k−1


(nk

)satisfy the





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 . . .



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




Pascal’s 4


)(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 . . .



Pascal’s 4


)(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




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




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?




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?




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?




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?




11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1




11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1


n∑k=0

(n

k

)= 2n




11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1


n∑k=0

(n

k

)= 2n




11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1


n∑i=k

(i

k

)=

(n + 1

k + 1

)




11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1


n∑i=k

(i

k

)=

(n + 1

k + 1

)




11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1


n∑k=0

(−1)k(n

k

)= 0




11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1


n∑k=0

(−1)k(n

k

)= 0




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




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



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

).




Theorem

For n ≥ 0,n∑

k=0

(n

k

)= 2n.





which is (n

0

)+

(n

1

)+ · · ·+

(n

n

).




Theorem

For n ≥ 0,n∑

k=0

(n

k

)= 2n.





which is (n

0

)+

(n

1

)+ · · ·+

(n

n

).




Theorem

For n ≥ 0,n∑

k=0

(n

k

)= 2n.





which is (n

0

)+

(n

1

)+ · · ·+

(n

n

).



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).




Theorem

For n ≥ 1,n∑

k=0

(−1)k(n

k

)= 0.








Theorem

For n ≥ 1,n∑

k=0

(−1)k(n

k

)= 0.








Theorem

For n ≥ 1,n∑

k=0

(−1)k(n

k

)= 0.








Theorem

For n ≥ 1,n∑

k=0

(−1)k(n

k

)= 0.








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|.






i. f (X ) = X ∪ {1} if 1 /∈ X , and

ii. f (X ) = X − {1} if 1 ∈ X .







i. f (X ) = X ∪ {1} if 1 /∈ X , and

ii. f (X ) = X − {1} if 1 ∈ X .







i. f (X ) = X ∪ {1} if 1 /∈ X , and

ii. f (X ) = X − {1} if 1 ∈ X .




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.




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.





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.




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.




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).




Let g(x) =∑n≥0

xn = 1 + x + x2 + x3 + · · · .








Let g(x) =∑n≥0

xn = 1 + x + x2 + x3 + · · · .








Let g(x) =∑n≥0

xn = 1 + x + x2 + x3 + · · · .







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.





1− x − x2?

F (x) = 1 + x + 2x2 + 3x3 + 5x4 + 8x5 + 13x6 + 21x7 + 34x8 + · · ·

=∑n≥0

fnxn.





1− x − x2?

F (x) = 1 + x + 2x2 + 3x3 + 5x4 + 8x5 + 13x6 + 21x7 + 34x8 + · · ·=

∑n≥0

fnxn.



The End

Thanks for listening.

Questions?



The End

Thanks for listening.Questions?


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Combinatorial Enumeration in Pascal’s Triangleramanujan.math.trinity.edu › bmiceli › research › MajorsS16.pdf · Combinatorial Enumeration in Pascal’s Triangle How To Count