New Discrete Mathematics and Its...

Discrete Mathematics and Its ApplicationsLecture 3: Counting: Pigeonhole Principle and Binomial Coefficients

MING GAO

DaSE@ ECNU(for course related communications)

[email protected]

Oct. 21, 2019

Outline

1 Pigeonhole Principle

2 Binomial Coefficient

3 Pascal’s Triangle

4 Take-aways

MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Oct. 21, 2019 2 / 33

Pigeonhole Principle

The sock problem

I have n pairs of socks. Each pair is different from the other pair.How many socks do I have to pick out to be sure that I have at leastone matching pair.

The answer of the previous question seems obvious. But it appears to bevery useful in numerous cases. It is called the pigeonhole principle.

The pigeonhole principle

If we put n+ 1 objects into n boxes, at least one box gets more thanone objects (also called Dirichlet drawer principle).

Example

Suppose that a flock of 20 pigeons flies into a set of 19 pigeonholes toroost. Because there are 20 pigeons but only 19 pigeonholes, a least oneof these 19 pigeonholes must have at least two pigeons in it.

The sock problem

Example

The sock problem

Example

The sock problem

Example

Students with the same birthday

It is quite often that you find people with the same birthday.

Since there is at most 366 days in a year, the pigeonholeprinciple states that if you have 367 people is a room, there isat least one pair with the same birthday.

But that’s the worst case scenario, as it is more common tofind people with the same birthday. (In the next class, we willtry to see if there is a pair of students in the class with thesame birthday.)

So, let’s think about the probability that there are twostudents with the same birthday in a room with 40 students.

A simple case

Let’s start with 2 people in the room.

Notes: While we have not defined properly what probabilities mean,

we can count the number of all possibilities and the number of cases

that we are interested in, and then calculate probability as the ratio

between the two. E.g., if there are 50 possible outcomes and 30 of

them are the ones we are interested in, the probability is 0.6. (Note

that we assume that every outcome is equally likely.)

How many possible birthdays can two people have?

Since each person has 366 choices, and the first person and thesecond person can choose independently, the number is366 · 366.

How many possible ways can they share the same birthday?

Since the first person has 366 choices, and the second personhas to choose the same day, there are only 366 ways.

Thus, the probability is 3663662 = 0.0027, very unlikely.

A simple case

between the two.

E.g., if there are 50 possible outcomes and 30 of

A simple case

them are the ones we are interested in, the probability is 0.6.

A simple case

3 people

Let’s consider 3 people.

How many possible birthdays can 3 people have?

Since each person has 366 choices, and each person can chooseindependently, the number is 366 · 366 · 366 = 3663.

How many possible ways can at least two of them share thesame birthday?

There are many cases.So let’s think about the case when everyone do not share anybirthdays.The first person has 366 choices. The second one has366− 1 = 365 choices. The third one has 366− 2 = 364choices. Thus, the number of ways they do not share anybirthdays is 366 · 365 · 364.Notice that this is the number of ordered subsets.

Thus, the probability that they do not share birthdays is366·365·364

3663 = 0.9918. Thus the probability that two of themshare a birthday is 1− 0.9918 = 0.0082.

3 people

There are many cases.

So let’s think about the case when everyone do not share anybirthdays.The first person has 366 choices. The second one has366− 1 = 365 choices. The third one has 366− 2 = 364choices. Thus, the number of ways they do not share anybirthdays is 366 · 365 · 364.Notice that this is the number of ordered subsets.

3 people

There are many cases.So let’s think about the case when everyone do not share anybirthdays.The first person has 366 choices.

The second one has366− 1 = 365 choices. The third one has 366− 2 = 364choices. Thus, the number of ways they do not share anybirthdays is 366 · 365 · 364.Notice that this is the number of ordered subsets.

3 people

There are many cases.So let’s think about the case when everyone do not share anybirthdays.The first person has 366 choices. The second one has366− 1 = 365 choices.

The third one has 366− 2 = 364choices. Thus, the number of ways they do not share anybirthdays is 366 · 365 · 364.Notice that this is the number of ordered subsets.

3 people

There are many cases.So let’s think about the case when everyone do not share anybirthdays.The first person has 366 choices. The second one has366− 1 = 365 choices. The third one has 366− 2 = 364choices.

Thus, the number of ways they do not share anybirthdays is 366 · 365 · 364.Notice that this is the number of ordered subsets.

3 people

There are many cases.So let’s think about the case when everyone do not share anybirthdays.The first person has 366 choices. The second one has366− 1 = 365 choices. The third one has 366− 2 = 364choices. Thus, the number of ways they do not share anybirthdays is 366 · 365 · 364.

Notice that this is the number of ordered subsets.

3 people

40 people

Let’s extend our previous argument to the case with 40 people.

36640.

How many possible ways that they do not sure any birthdays?

This is the number of ordered subsets with 40 elements of a366-set.Thus, there are 366 · 365 · 364 · · · 327 ways.

Thus, the probability that they do not share birthdays is

366 · 365 · 364 · · · 327

36640.

Umm... how small is it?

Again you can use a computer to compute the exact value ofthis quantity. For example, you may want to use WolframAlpha.

Anyway, we will try to estimate it using basic mathematicaltools.

40 people

36640.

366 · 365 · 364 · · · 327

36640.

40 people

36640.

366 · 365 · 364 · · · 327

36640.

40 people

36640.

This is the number of ordered subsets with 40 elements of a366-set.

Thus, there are 366 · 365 · 364 · · · 327 ways.

366 · 365 · 364 · · · 327

36640.

40 people

36640.

366 · 365 · 364 · · · 327

36640.

40 people

36640.

366 · 365 · 364 · · · 327

36640.

40 people

36640.

366 · 365 · 364 · · · 327

36640.

40 people

36640.

366 · 365 · 364 · · · 327

36640.

40 people

36640.

366 · 365 · 364 · · · 327

36640.

General case: n days k people

Let’s continue on the general case. When we have k peopleand a year contains n days, the probability that no two peopleshare the same birthday is

n · (n − 1) · (n − 2) · · · (n − k + 1)

If this number is very close to 0, then it is very unlikely that notwo people share the same birthday, i.e., it is very likely thatthere exists two people with the same birthday.

General case: n days k people

Let’s continue on the general case. When we have k peopleand a year contains n days, the probability that no two peopleshare the same birthday is

n · (n − 1) · (n − 2) · · · (n − k + 1)

If this number is very close to 0, then it is very unlikely that notwo people share the same birthday, i.e., it is very likely thatthere exists two people with the same birthday.

A few tweaks

Dealing with small numbers is sometimes troublesome. (Thereason will be more apparent later when we start introducingthe tools.) So let’s consider the reciprocal instead:

n · (n − 1) · (n − 2) · · · (n − k + 1).

The top term looks easy to deal with; the bottom one doesnot. Let’s break up the product:(n

n − 1

n − 2

)· · ·(

n − k + 1

If you look closely at this product, you can see that each termis at least one. In the beginning, the terms are very close toone and they get larger at the end.

A few tweaks

n · (n − 1) · (n − 2) · · · (n − k + 1).

n − 1

n − 2

)· · ·(

n − k + 1

If you look closely at this product, you can see that each termis at least one. In the beginning, the terms are very close toone and they get larger at the end.

A few tweaks

n · (n − 1) · (n − 2) · · · (n − k + 1).

n − 1

n − 2

)· · ·(

n − k + 1

If you look closely at this product, you can see that each termis at least one. In the beginning, the terms are very close toone and they get larger at the end.MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Oct. 21, 2019 9 / 33

The logarithms

There is a nice tool that you can turn multiplications toadditions: logarithms. So let’s try to take the logarithms;

n − 1

n − 2

)· · ·(

n − k + 1

= ln(nn

n − 1

n − 2

)+ · · ·+ ln

n − k + 1

The terms do not look that much better. But there’s a nicefact about the natural logarithms.

The logarithms

There is a nice tool that you can turn multiplications toadditions: logarithms. So let’s try to take the logarithms; weget

n − 1

n − 2

)· · ·(

n − k + 1

= ln(nn

n − 1

n − 2

)+ · · ·+ ln

n − k + 1

The logarithms

There is a nice tool that you can turn multiplications toadditions: logarithms. So let’s try to take the logarithms; weget

n − 1

n − 2

)· · ·(

n − k + 1

= ln(nn

n − 1

n − 2

)+ · · ·+ ln

n − k + 1

ln x : the upper bound

Fact:ln x ≤ x − 1

This fact can be proved with elementary calculus. But it is fairly clear ifyou plot the functions ln x and x − 1.

So let’s do that at Wolfram Alpha.

ln x : the upper bound

Fact:ln x ≤ x − 1

This fact can be proved with elementary calculus. But it is fairly clear ifyou plot the functions ln x and x − 1.So let’s do that at Wolfram Alpha.

ln x : the lower bound

We know thatln x ≤ x − 1

If we use the fact that ln 1x = − ln x , we can obtain the lower bound.

ln x = − ln1

x≥ −

x− 1

x − 1

Let’s conclude by stating the lemma:

x − 1

x≤ ln x ≤ x − 1.

ln x : the lower bound

We know thatln x ≤ x − 1

If we use the fact that ln 1x = − ln x , we can obtain the lower bound.

ln x = − ln1

x≥ −

x− 1

x − 1

Let’s conclude by stating the lemma:

x − 1

x≤ ln x ≤ x − 1.

The lower bound

Let’s look at each term in the sum: ln(

nn−j

). Using the lower bound in

Lemma 1, we get that

n − j

nn−j − 1

nn−j

n−n+jn−jn

n − 1

n − 2

)· · ·(

n − k + 1

= ln(nn

n − 1

n − 2

)+ · · ·+ ln

n − k + 1

)≥ 0

n+ · · ·+ k − 1

n(1 + 2 + · · ·+ (k − 1)) =

k(k − 1)

The lower bound

Let’s look at each term in the sum: ln(

nn−j

). Using the lower bound in

Lemma 1, we get that

n − j

nn−j − 1

nn−j

n−n+jn−jn

n − 1

n − 2

)· · ·(

n − k + 1

= ln(nn

n − 1

n − 2

)+ · · ·+ ln

n − k + 1

)≥ 0

n+ · · ·+ k − 1

n(1 + 2 + · · ·+ (k − 1)) =

k(k − 1)

The upper bound

Again, let’s look at each term in the sum: ln(

nn−j

). Using the upper

bound in Lemma 1, we get that

n − j

)≤ n

n − j− 1 =

n − j.

n − 1

n − 2

)· · ·(

n − k + 1

n − 0+

n − 1+

n − 2+ · · ·+ k − 1

n − k + 1

n − k + 1+

n − k + 1+ · · ·+ k − 1

n − k + 1

n − k + 1(1 + 2 + · · ·+ (k − 1)) =

k(k − 1)

2(n − k + 1).

The upper bound

Again, let’s look at each term in the sum: ln(

nn−j

). Using the upper

bound in Lemma 1, we get that

n − j

)≤ n

n − j− 1 =

n − j.

n − 1

n − 2

)· · ·(

n − k + 1

n − 0+

n − 1+

n − 2+ · · ·+ k − 1

n − k + 1

n − k + 1+

n − k + 1+ · · ·+ k − 1

n − k + 1

n − k + 1(1 + 2 + · · ·+ (k − 1)) =

k(k − 1)

2(n − k + 1).

Using the derived upper and lower bounds, we get

ek(k−1)

2n ≤ nk

n(n − 1)(n − 2) · · · (n − k + 1)≤ e

k(k−1)2(n−k+1)

Let’s plug in n = 366 and k = 40:

8.42 ≤ 36640

366 · 365 · · · 327≤ 10.86.

So the probability that we get no two people with the same birthday isbetween 1/8.42 ≈ 0.118 and 1/10.86 ≈ 0.092. So we have high chance offinding two students with the same birthday. This is pretty close as theactual value is 0.1094.

ek(k−1)

2n ≤ nk

n(n − 1)(n − 2) · · · (n − k + 1)≤ e

k(k−1)2(n−k+1)

8.42 ≤ 36640

366 · 365 · · · 327≤ 10.86.

ek(k−1)

2n ≤ nk

n(n − 1)(n − 2) · · · (n − k + 1)≤ e

k(k−1)2(n−k+1)

8.42 ≤ 36640

366 · 365 · · · 327≤ 10.86.

So the probability that we get no two people with the same birthday isbetween 1/8.42 ≈ 0.118 and 1/10.86 ≈ 0.092. So we have high chance offinding two students with the same birthday.

This is pretty close as theactual value is 0.1094.

ek(k−1)

2n ≤ nk

n(n − 1)(n − 2) · · · (n − k + 1)≤ e

k(k−1)2(n−k+1)

8.42 ≤ 36640

366 · 365 · · · 327≤ 10.86.

The generalized pigeonhole principle

Corollary: A function f from a set with k + 1 or more elements toa set with k elements is not one-to-one.

Theorem: If N objects are placed into k boxes, then there is at leastone box containing at least dNk e objects.

Proof.

Suppose that none of the boxes contains more than dNk e − 1objects. Then, the total number of objects is at most

k(dNke − 1) < k((

k+ 1)− 1) = N,

where the inequality dNk e <Nk + 1 has been used.

This is a contradiction because there are a total of N objects.

Proof.

Suppose that none of the boxes contains more than dNk e − 1objects.

Then, the total number of objects is at most

k(dNke − 1) < k((

k+ 1)− 1) = N,

Proof.

k(dNke − 1) < k((

k+ 1)− 1) = N,

Proof.

k(dNke − 1) < k((

k+ 1)− 1) = N,

Some elegant applications of the Pigeonhole principle I

Question: During a month with 30 days, a baseball team plays atleast one game a day, but no more than 45 games. Show that theremust be a period of some number of consecutive days during whichthe team must play exactly 14 games.

Solution

Let aj be the number of games played on or before the j−th day ofthe month. Then 1 ≤ a1 ≤ a2 ≤ · · · ≤ a30 ≤ 45.

Moreover, 15 ≤ ai + 14 ≤ 59 is also an increasing sequence ofdistinct positive integers. The 60 positive integers have1 ≤ a1, a2, · · · , a30, a1 + 14, a2 + 14, · · · , a30 + 14 ≤ 59.Hence, by the pigeonhole principle two of these integers are equal.Because aj are all distinct and aj + 14 are all distinct, there must beindices i and j with ai = aj + 14. This means that exactly 14 gameswere played from day j + 1 to day i .

Solution

Let aj be the number of games played on or before the j−th day ofthe month. Then 1 ≤ a1 ≤ a2 ≤ · · · ≤ a30 ≤ 45.Moreover, 15 ≤ ai + 14 ≤ 59 is also an increasing sequence ofdistinct positive integers. The 60 positive integers have1 ≤ a1, a2, · · · , a30, a1 + 14, a2 + 14, · · · , a30 + 14 ≤ 59.

Hence, by the pigeonhole principle two of these integers are equal.Because aj are all distinct and aj + 14 are all distinct, there must beindices i and j with ai = aj + 14. This means that exactly 14 gameswere played from day j + 1 to day i .

Solution

Let aj be the number of games played on or before the j−th day ofthe month. Then 1 ≤ a1 ≤ a2 ≤ · · · ≤ a30 ≤ 45.Moreover, 15 ≤ ai + 14 ≤ 59 is also an increasing sequence ofdistinct positive integers. The 60 positive integers have1 ≤ a1, a2, · · · , a30, a1 + 14, a2 + 14, · · · , a30 + 14 ≤ 59.Hence, by the pigeonhole principle two of these integers are equal.Because aj are all distinct and aj + 14 are all distinct, there must beindices i and j with ai = aj + 14. This means that exactly 14 gameswere played from day j + 1 to day i .

Some elegant applications of the Pigeonhole principle II

Question: Show that among any n+1 positive integers not exceeding2n there must be an integer that divides one of the other integers.

Solution

Write each of the n + 1 integers a1, a2, · · · , an+1 as a power of 2times an odd integer. In other words, let aj = 2kjqj forj = 1, 2, · · · , n + 1, where kj is a nonnegative integer and qj is odd.Integers q1, q2, · · · , qn+1 are all odd positive integers less than 2n.Because there are only n odd positive integers less than 2n, itfollows from the pigeonhole principle that two of integersq1, q2, · · · , qn+1 must be equal. Therefore, there are distinctintegers i and j such that qi = qj . Let q be the common value ofqi and qj . Then, ai = 2kiq and aj = 2kjq. It follows that if ki < kj, then ai divides aj ; while if ki > kj , then aj divides ai .

Ramsey theory

Theorem: Every sequence of n2 + 1 distinct real numbers contains asubsequence of length n + 1 that is increasing or decreasing strictly.Proof: Let ai be a sequence of n2 + 1 distinct real numbers. As-sociate (ik , dk) to term ak , where ik and dk are the lengths of thelongest increasing and longest decreasing subsequences starting atak , respectively.

Suppose that there are no increasing or decreas-ing subsequences of length n + 1. Then ik ≤ n and dk ≤ n, fork = 1, 2, · · · , n2 + 1. Hence, by the product rule there are n2 pos-sible ordered pairs for (ik , dk). By the pigeonhole principle, two ofthese n2 + 1 ordered pairs are equal, i.e., there exist terms as andat (s < t) s.t. is = it and ds = dt . Because the terms are distinct,either as < at or as > at . If as < at , then, because is = it , anincreasing subsequence of length it + 1 can be built starting at as , bytaking as followed by an increasing subsequence of length it begin-ning at at . This is a contradiction. Similarly, if as > at , the samereasoning shows that ds > dt , which is a contradiction.

Ramsey theory

Theorem: Every sequence of n2 + 1 distinct real numbers contains asubsequence of length n + 1 that is increasing or decreasing strictly.Proof: Let ai be a sequence of n2 + 1 distinct real numbers. As-sociate (ik , dk) to term ak , where ik and dk are the lengths of thelongest increasing and longest decreasing subsequences starting atak , respectively. Suppose that there are no increasing or decreas-ing subsequences of length n + 1. Then ik ≤ n and dk ≤ n, fork = 1, 2, · · · , n2 + 1.

Hence, by the product rule there are n2 pos-sible ordered pairs for (ik , dk). By the pigeonhole principle, two ofthese n2 + 1 ordered pairs are equal, i.e., there exist terms as andat (s < t) s.t. is = it and ds = dt . Because the terms are distinct,either as < at or as > at . If as < at , then, because is = it , anincreasing subsequence of length it + 1 can be built starting at as , bytaking as followed by an increasing subsequence of length it begin-ning at at . This is a contradiction. Similarly, if as > at , the samereasoning shows that ds > dt , which is a contradiction.

Ramsey theory

Theorem: Every sequence of n2 + 1 distinct real numbers contains asubsequence of length n + 1 that is increasing or decreasing strictly.Proof: Let ai be a sequence of n2 + 1 distinct real numbers. As-sociate (ik , dk) to term ak , where ik and dk are the lengths of thelongest increasing and longest decreasing subsequences starting atak , respectively. Suppose that there are no increasing or decreas-ing subsequences of length n + 1. Then ik ≤ n and dk ≤ n, fork = 1, 2, · · · , n2 + 1. Hence, by the product rule there are n2 pos-sible ordered pairs for (ik , dk). By the pigeonhole principle, two ofthese n2 + 1 ordered pairs are equal, i.e., there exist terms as andat (s < t) s.t. is = it and ds = dt .

Because the terms are distinct,either as < at or as > at . If as < at , then, because is = it , anincreasing subsequence of length it + 1 can be built starting at as , bytaking as followed by an increasing subsequence of length it begin-ning at at . This is a contradiction. Similarly, if as > at , the samereasoning shows that ds > dt , which is a contradiction.

Ramsey theory

Theorem: Every sequence of n2 + 1 distinct real numbers contains asubsequence of length n + 1 that is increasing or decreasing strictly.Proof: Let ai be a sequence of n2 + 1 distinct real numbers. As-sociate (ik , dk) to term ak , where ik and dk are the lengths of thelongest increasing and longest decreasing subsequences starting atak , respectively. Suppose that there are no increasing or decreas-ing subsequences of length n + 1. Then ik ≤ n and dk ≤ n, fork = 1, 2, · · · , n2 + 1. Hence, by the product rule there are n2 pos-sible ordered pairs for (ik , dk). By the pigeonhole principle, two ofthese n2 + 1 ordered pairs are equal, i.e., there exist terms as andat (s < t) s.t. is = it and ds = dt . Because the terms are distinct,either as < at or as > at . If as < at , then, because is = it , anincreasing subsequence of length it + 1 can be built starting at as , bytaking as followed by an increasing subsequence of length it begin-ning at at . This is a contradiction.

Similarly, if as > at , the samereasoning shows that ds > dt , which is a contradiction.

Ramsey theory

Theorem: Every sequence of n2 + 1 distinct real numbers contains asubsequence of length n + 1 that is increasing or decreasing strictly.Proof: Let ai be a sequence of n2 + 1 distinct real numbers. As-sociate (ik , dk) to term ak , where ik and dk are the lengths of thelongest increasing and longest decreasing subsequences starting atak , respectively. Suppose that there are no increasing or decreas-ing subsequences of length n + 1. Then ik ≤ n and dk ≤ n, fork = 1, 2, · · · , n2 + 1. Hence, by the product rule there are n2 pos-sible ordered pairs for (ik , dk). By the pigeonhole principle, two ofthese n2 + 1 ordered pairs are equal, i.e., there exist terms as andat (s < t) s.t. is = it and ds = dt . Because the terms are distinct,either as < at or as > at . If as < at , then, because is = it , anincreasing subsequence of length it + 1 can be built starting at as , bytaking as followed by an increasing subsequence of length it begin-ning at at . This is a contradiction. Similarly, if as > at , the samereasoning shows that ds > dt , which is a contradiction.

Binomial Coefficient

The binomial coefficients

There is a reason why the term(nk

)is called the binomial coefficients. In

this lecture, we will discuss

the Pascal’s triangle,

the binomial theorem, and

advanced counting with binomial coefficients.

Polynomial expansions

Let’s start by looking at polynomial of the form (x + y)n. Let’s start withsmall values of n:

(x + y)1 = x + y

(x + y)2 = x2 + 2 · xy + y2

(x + y)3 = x3 + 3 · x2y + 3 · xy2 + y3

(x + y)4 = x4 + 4 · x3y + 6 · x2y2 + 4 · xy3 + y4.

Let’s focus on the coefficient of each term. You may notice that terms xn

and yn always have 1 as their coefficients. Why is that?Let’s look further at the coefficients of terms xn−1y . Do you see anypattern in their coefficients? Can you explain why?

Polynomial expansions

Let’s start by looking at polynomial of the form (x + y)n. Let’s start withsmall values of n:

(x + y)1 = x + y

(x + y)2 = x2 + 2 · xy + y2

(x + y)3 = x3 + 3 · x2y + 3 · xy2 + y3

(x + y)4 = x4 + 4 · x3y + 6 · x2y2 + 4 · xy3 + y4.

Let’s focus on the coefficient of each term. You may notice that terms xn

and yn always have 1 as their coefficients. Why is that?Let’s look further at the coefficients of terms xn−1y . Do you see anypattern in their coefficients? Can you explain why?

Another way to look at it

Let’s take a look at (x + y)4 again. It is

(x + y)(x + y)(x + y)(x + y).

How do we get x4 in the expansion?

For every factory, you have topick x .

How do we get x3y in the expansion? Out of the 4 factors, you haveto pick y in one of the factor (or you have to pick x in 3 of thefactors). Thus there are

)ways to do so.

(x + y)(x + y)(x + y)(x + y).

How do we get x4 in the expansion? For every factory, you have topick x .

How do we get x3y in the expansion?

Out of the 4 factors, you haveto pick y in one of the factor (or you have to pick x in 3 of thefactors). Thus there are

)ways to do so.

(x + y)(x + y)(x + y)(x + y).

How do we get x3y in the expansion? Out of the 4 factors, you haveto pick y in one of the factor (or you have to pick x in 3 of thefactors).

Thus there are(4

)ways to do so.

(x + y)(x + y)(x + y)(x + y).

How do we get x3y in the expansion? Out of the 4 factors, you haveto pick y in one of the factor (or you have to pick x in 3 of thefactors). Thus there are

)ways to do so.

The binomial theorem

Theorem

If you expand (x + y)n, the coefficient of the term xkyn−k is(nk

That is,

(x + y)n =n∑

)xkyn−k =

n − 1

)xn−1y1 +

n − 2

)xn−2y2 + · · ·+

)xyn−1 +

Additional applications of the binomial theorem

The binomial theorem can be used to prove various identities regarding thebinomial coefficients.

Corollary 1

Let n be a nonnegative integer, then(n

)+ · · ·+

n − 1

)= 2n[Hint: (1 + 1)n = 2n].

Quick check. Can you prove that(n

)−(n

)+ · · · = 0.

Note that this statements says that the number of odd subsets equalsthe number of even subsets.

Additional applications of the binomial theorem

The binomial theorem can be used to prove various identities regarding thebinomial coefficients.

Corollary 1

Let n be a nonnegative integer, then(n

)+ · · ·+

n − 1

)= 2n[Hint: (1 + 1)n = 2n].

Quick check. Can you prove that(n

)−(n

)+ · · · = 0.

Note that this statements says that the number of odd subsets equalsthe number of even subsets.

Corollary

Corollary 2

Let n be a nonnegative integer. Then(n

)−(n

)+ · · · = 0.[Hint: (1− 1)n = 0]

Corollary 3

Let n be a nonnegative integer. Then

n∑k=0

)= 3n[Hint: (1 + 2)n = 3n].

Corollary

Corollary 2

Let n be a nonnegative integer. Then(n

)−(n

)+ · · · = 0.[Hint: (1− 1)n = 0]

Corollary 3

Let n be a nonnegative integer. Then

n∑k=0

)= 3n[Hint: (1 + 2)n = 3n].

Pascal’s Triangle

Triangle binomial coefficient

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

Pascal’s Triangle

The Triangle

If we move the numbers in the table slightly to the right, the tablebecomes the Pascal’s triangle.

1 2 11 3 3 1

1 4 6 4 11 5 10 10 5 1

1 6 15 20 15 6 1...

......

The table and the binomial coefficients have many other interestingproperties.

Pascal’s Triangle

The Triangle

If we move the numbers in the table slightly to the right, the tablebecomes the Pascal’s triangle.

1 2 11 3 3 1

1 4 6 4 11 5 10 10 5 1

1 6 15 20 15 6 1...

......

The table and the binomial coefficients have many other interestingproperties.

Pascal’s Triangle

Pascal’s triangle

)Yanghui’s triangle Pascal’s triangle

Pascal’s triangle wasknown in the early11th century throughthe work of Chinesemathematicians Jia X-ian (1010-1070) andYang Hui (1238-1298).

Pascal’s Triangle

Next observation I

Pascal’s identity

Theorem: Let n and k be positive integers with n ≥ k . Then(n + 1

k − 1

Vandenmonde’s identity

Theorem: Let m, n and r be nonnegative integers with r not exceed-ing either m or n. Then(

r∑k=0

r − k

Pascal’s Triangle

Next observation I

Pascal’s identity

Theorem: Let n and k be positive integers with n ≥ k . Then(n + 1

k − 1

Vandenmonde’s identity

Theorem: Let m, n and r be nonnegative integers with r not exceed-ing either m or n. Then(

r∑k=0

r − k

Pascal’s Triangle

Next observation II

1 2 11 3 3 1

1 4 6 4 11 5 10 10 5 1

1 6 15 20 15 6 11 7 21 35 35 21 7 1

Let’s try to compute the sum of squares of numbers in each row.

12 = 1

12 + 12 = 2

12 + 22 + 12 = 6

12 + 32 + 32 + 12 = 20

12 + 42 + 62 + 42 + 12 = 70

Pascal’s Triangle

Next observation II

1 2 11 3 3 1

1 4 6 4 11 5 10 10 5 1

1 6 15 20 15 6 11 7 21 35 35 21 7 1

12 = 1

12 + 12 = 2

12 + 22 + 12 = 6

12 + 32 + 32 + 12 = 20

12 + 42 + 62 + 42 + 12 = 70

Pascal’s Triangle

Next observation II

1 2 11 3 3 1

1 4 6 4 11 5 10 10 5 1

1 6 15 20 15 6 11 7 21 35 35 21 7 1

12 = 1

12 + 12 = 2

12 + 22 + 12 = 6

12 + 32 + 32 + 12 = 20

12 + 42 + 62 + 42 + 12 = 70

Pascal’s Triangle

Next observation II

1 2 11 3 3 1

1 4 6 4 11 5 10 10 5 1

1 6 15 20 15 6 11 7 21 35 35 21 7 1

12 = 1

12 + 12 = 2

12 + 22 + 12 = 6

12 + 32 + 32 + 12 = 20

12 + 42 + 62 + 42 + 12 = 70

Pascal’s Triangle

Next observation II

1 2 11 3 3 1

1 4 6 4 11 5 10 10 5 1

1 6 15 20 15 6 11 7 21 35 35 21 7 1

12 = 1

12 + 12 = 2

12 + 22 + 12 = 6

12 + 32 + 32 + 12 = 20

12 + 42 + 62 + 42 + 12 = 70

Pascal’s Triangle

Theorem: Let n be a nonnegative integer. Then(2n

n∑k=0

Proof.n∑

n − k

Pascal’s Triangle

Corollary 4: Let n and r be a nonnegative integers with r ≤ n. Then(n + 1

n∑k=r

Proof.

Let(n+1r+1

)counts the bit strings of length n + 1 containing r + 1

ones.Ways Location of last 1 Counting

Way 1: r + 1(rr

)Way 2: r + 2

)Way 3: r + 3

)· · · · · ·Way n − r + 1: n + 1

Pascal’s Triangle

Corollary 4: Let n and r be a nonnegative integers with r ≤ n. Then(n + 1

n∑k=r

Proof.

Let(n+1r+1

)counts the bit strings of length n + 1 containing r + 1

ones.Ways Location of last 1 CountingWay 1: r + 1

)Way 2: r + 2

)Way 3: r + 3

)· · · · · ·Way n − r + 1: n + 1

Take-aways

Conclusions

Pascal’s Triangle

New Discrete Mathematics and Its...

Documents

Transcript of New Discrete Mathematics and Its...