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Number Theory and Counting Method

Divisors

-Least common divisor

-Greatest common multiple

DivisorsDefinition

𝑛 and 𝑑 are integers 𝑑 ≠ 0

𝑑 divides 𝑛 if there exists 𝑞 satisfying 𝑛 = 𝑑𝑞𝑞 ∶ the quotient, 𝑑 ∶ the divisor, 𝑛 ∶ the number

If 𝑑 divides 𝑛 : 𝑑|𝑛

𝑑 does not divides 𝑛 : 𝑑 ∤ 𝑛

Ex. 21 = 3.73|21, 3 is a divisor (factor) of 21

There exist an integer 𝑞 such that

𝑛 = 𝑑𝑞, 𝑑, 𝑛 ∈ 𝐼+, 1 ≤ 𝑞Therefore, 𝑑 ≤ 𝑑𝑞 = 𝑛.

TheoremLet 𝑚, 𝑛 and 𝑑 be integers

(a) if 𝑑|𝑚 and 𝑑|𝑛 then𝑑|(𝑚 + 𝑛)

(b) if 𝑑|𝑚 and 𝑑|𝑛 then𝑑|(𝑚 − 𝑛)

(c) if 𝑑|𝑚 then 𝑑|𝑚𝑛

proof

(a) Suppose d|m and d|n by definition𝑚 = 𝑑𝑞1

and 𝑛 = 𝑑𝑞2

Thus (𝑚 + 𝑛) = 𝑑(𝑞1 + 𝑞2)

Therefor

There exist 𝑞1 + 𝑞2 is the quotient of it, (a) is proof

Prime and Composite number

• An integer greater than 1 whose only positive divisors are itself and 1 is called prime.

• An integer greater than 1 that is not prime is called composite

Ex.

23 is prime

34 is composite because it is divided by 17

If 𝑛 > 1 is composite, then ∃𝑑, 1 < 𝑑 < 𝑛

To test use 2,3,… , 𝑛 − 1

If no integer in this list divides 𝑛, then 𝑛 is prime

Composite number theorem (cont.)

A positive integer 𝑛 greater than 1 is composite if and only if 𝑛 has a divisor 𝑑 satisfying 2 ≤ 𝑑 ≤ 𝑛

Proof

- If 𝑛 is composite, then 𝑛 has a divisor 𝑑 satisfying 2 ≤𝑑 ≤ 𝑛

- If 𝑛 has a divisor 𝑑 satisfying 2 ≤ 𝑑 ≤ 𝑛 then 𝑛 is composite

Ex1 Determine whether 43 is 𝑝𝑟𝑖𝑚𝑒

Try 2,3,… , 21 = 43

None of these number divides 43 agree with the condition 𝑛 mod 𝑑 = 0

∴ 43 is 𝑝𝑟𝑖𝑚𝑒

Ex2. Determine 451

Try 2,3,… , 21 = ⌊ 451⌋

It was founded that 11 is agree with the condition 𝑛 mod 𝑑 = 0

∴ 451 is 𝑐𝑜𝑚𝑝𝑜𝑠𝑖𝑡𝑒

Fundamental Theorem of Arithmetic

• Any integer greater than 1 can be written as a product of primes. Moreover, if the primes are written in non decreasing order the factorization is unique.

• The number of primes is infinite

Ex. Produce a prime larger than 112,3,5,7,11 are 𝑝𝑟𝑖𝑚𝑒𝑠

Let𝑚 = 2 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 11 + 1 = 2311

2311 is 𝑝𝑟𝑖𝑚𝑒

The greatest common divisor (GCD)

Definition

Let 𝑚 and 𝑛 be integers with not both 𝑚 and 𝑛 zero. The gdc is an integer that divides both 𝑚 and 𝑛.

Denote 𝑔𝑑𝑐(𝑚, 𝑛)

Ex. The positive divisor of 30 are1, 2, 3, 5, 6, 10, 15, 30

The positive divisor of 105 are1, 3, 5, 7, 15, 21, 35, 105

The common positive1, 3, 5, 15

Thus𝑔𝑑𝑐 30,105 = 15

It can be also find by looking carefully at their prime factorization30 = 2 ∙ 3 ∙ 5, 105 = 3 ∙ 5 ∙ 7

3 and 5 are the prime factorization of 30 and 105 thus 𝑔𝑑𝑐(30,105) = 3 ∙ 5 = 15

Ex. The positive divisors of 30 are1, 2, 3, 5, 6, 10, 15, 30

The positive divisors of 105 are1, 3, 5, 7, 15, 21, 35, 105

The common divisors of 30 and 105 are1,3,5,15

And gcd 30,105 = 15

Ex2. Find the gcd by looking at their prime factorization30 = 2 ⋅ 3 ⋅ 5 105 = 3 ⋅ 5 ⋅ 7

The common prime factorization for 30 and 105 are 3 and 5

Thus gcd(30,105) = 3 ⋅ 5 = 15

GDC with prime factorization

Theorem

Let 𝑚 and 𝑛 be integers, 𝑚 > 1, 𝑛 > 1 with prime factorization𝑚 = 𝑝1

𝑎1 ∙ 𝑝2𝑎2 ⋯𝑝𝑛

𝑎𝑛

and 𝑛 = 𝑝1

𝑏1 ∙ 𝑝2𝑏2 ⋯𝑝𝑛

𝑏𝑛

If the prime 𝑝𝑖 is not a factor of m, we let 𝑎𝑖 = 0. Similarly, if the prime 𝑝𝑖 is not a factor of 𝑛, we let 𝑏𝑖 = 0 Then

gcd 𝑚, 𝑛 = 𝑝1min 𝑎1,𝑏1 ∙ 𝑝2

min 𝑎2,𝑏2 ⋯𝑝𝑛min 𝑎𝑛,𝑏𝑛

Ex. gcd(82320, 950796)82320 = 24 ∙ 31 ∙ 51 ∙ 73 ∙ 110

950796 = 22 ∙ 31 ∙ 50 ∙ 74 ∙ 111

gcd 82320, 950796 = 2min 4,2 ∙ 3min 1,1 ∙ 5min 1,0 ∙ 7min 3,4 ∙ 11min 0,1

= 22 ∙ 31 ∙ 50 ∙ 73 ∙ 110

= 4116

The least common multiple

Definition

Let 𝑚 and 𝑛 be positive integer. A common multiple of 𝑚 and 𝑛 is an integer that is divisible by both 𝑚 and 𝑛, called the least common multiple.

𝑙𝑐𝑚(𝑚, 𝑛)

Ex. lcm(30,105) = 210

210 can be divided by both 30 and 105 and it is the least value30 = 2.3.5, 105 = 3.5.7

The lcm must contain 2, 3, 5, 7 thus lcm 30,105 = 2 ∙ 3 ∙ 5 ∙ 7 = 210

Also find lcm() by looking at their prime factorization30 = 2 ⋅ 3 ⋅ 5, 105 = 3 ⋅ 5 ⋅ 7

𝑙𝑐𝑚 30,105 = 2 ⋅ 3 ⋅ 5 ⋅ 7 = 210

LCM wit prime factorization

Let 𝑚 and 𝑛 be integers 𝑚 > 1, 𝑛 > 1 with prime factorizations𝑚 = 𝑝1

𝑎1𝑝2𝑎2 ⋯𝑝𝑛

𝑎𝑛

and 𝑛 = 𝑝1

𝑏1𝑝2𝑏2 ⋯𝑝𝑛

𝑏𝑛

If the prime 𝑝𝑖 is not a factor of 𝑚, we let 𝑎𝑖 = 0, Similarly, if the prime 𝑝𝑖 is not a factor of 𝑛, we let 𝑏𝑖 = 0, then

𝑙𝑐𝑚 𝑚, 𝑛 = 𝑝1max 𝑎1,𝑏1 𝑝2

max 𝑎2,𝑏2 ⋯𝑝𝑛max 𝑎𝑛,𝑏𝑛

Ex. lcm(82320, 950796)82320 = 24 ⋅ 31 ⋅ 51 ⋅ 73 ⋅ 110

950796 = 22 ⋅ 32 ⋅ 50 ⋅ 74 ⋅ 111

𝑙𝑐𝑚 82320, 950796 = 2max 4,2 ⋅ 3max 1,2 ⋅ 5max 1,0 ⋅ 7max 3,4 ⋅ 11max 0,1

= 22 ⋅ 32 ⋅ 51 ⋅ 74 ⋅ 111 = 19015920

Note

The product of gcd and lcm is equal to the product of the pair of numbersgcd 30,105 ⋅ 𝑙𝑐𝑚 30,105 = 30 ⋅ 105 = 3150 = 15 ⋅ 210

Representations of Integers and integer algorithm

For base 2

bit : a binary digit, value either 0 or 1

For Decimal number3854 = 3 ⋅ 103 + 8 ⋅ 102 + 5 ⋅ 101 + 4 ⋅ 100

Representation of integer

let n= integer that need to represent by a serie of binary number

𝑛 = 1 ⋅ 2𝑘 + 𝑏𝑘−12𝑘−1 +⋯+ 𝑏02

0

then2𝑘 ≤ 𝑛

and𝑛 = 1 ⋅ 2𝑘 + 𝑏𝑘−12

𝑘−1 +⋯+ 𝑏020

≤ 1 ⋅ 2𝑘 + 1 ⋅ 2𝑘−1 +⋯+ 1 ⋅ 20 = 2𝑘+1 − 1< 2𝑘+1

Counting Methods

Basic principle

Multiplication principle

If an activity can be constructed in 1 successive steps and step 1 can be done in 𝑛1 ways, step 2 can then be done in 𝑛2 ways,…, and step t can then be done in 𝑛𝑡 ways, then the number of different possible activities is

𝑛1 ⋅ 𝑛2⋯𝑛𝑡Addition principle

Suppose that 𝑋1, … , 𝑋𝑡 are sets and that the ith set 𝑋𝑖 has 𝑛𝑖elements. If {𝑋1, … , 𝑋𝑡} is a pairwise disjoint family (i.e., if 𝑖 ≠ 𝑗, 𝑋𝑖 ∩ 𝑋𝑗 = ∅),

the number of possible element that can be selected from 𝑋1or 𝑋2 or … or 𝑋𝑡 is

𝑛1 + 𝑛2 +⋯+ 𝑛𝑡

How many menu sets can be select from each categories: main course, appetizers, beverages

Solution : 24 possible outcome

NHT,NHM,NHC,NHR,NCT,NCM,NCC,NCR,

NFT,NFM,NFC,NFR,SHT,SHM,SHC,SHR,

SCT,SCM,SCC,SCR,SFT,SFM,SFC,SFR

Ex1

(a) How many strings of length 4 can be formed using the letters ABCDE if repetitions are not allowed ?

5 ⋅ 4 ⋅ 3 ⋅ 2 = 120

(b) How many strings of part (a) begin with the letter B?1 ⋅ 4 ⋅ 3 ⋅ 2 = 24

(c) How many strings of part (a) do not begin with the letter B ?120 − 24 = 96

Ex2

In a digital picture, we wish to encode the amount of light at each point as an eight-bit string. How many values are possible at one point?

2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 = 28 = 256

How many eight-bit strings begin either 101 or 111?

Bit 8,7,6 are tied together to form a bit and has 2 option 101 or 111

The 4 successive bit has the possible outcome 24

Thus2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 = 64

Permutations and Combinations

A permutation of n distinct elements 𝑥1, … , 𝑥𝑛 is an ordering of the 𝑛 elements

𝑥1⋯𝑥𝑛Theorem :

There are n! permutations of n elements𝑛 𝑛 − 1 …2 ⋅ 1 = 𝑛!

Ex. How many permutations of the letters 𝐴𝐵𝐶𝐷𝐸𝐹 contain substring 𝐷𝐸𝐹 ?

4!

Ex. How many ways can 6 person be seated around a circular table?

(6 − 1)! = 5! = 120

r-permutation

An r-permutation of n (distinct) elements 𝑥1, 𝑥2, … , 𝑥𝑛 is an ordering of an r-element subset of 𝑥1, 𝑥2, … , 𝑥𝑛 . The number of r-permutations of a set of n distinct elements is denoted 𝑃 𝑛, 𝑟

𝑃 𝑛, 𝑟 = 𝑛 𝑛 − 1 𝑛 − 2 … 𝑛 − 𝑟 + 1 , 𝑟 ≤ 𝑛

=𝑛!

𝑛 − 𝑟 !

Ex. 𝑋 = {𝑎, 𝑏, 𝑐}, How many 2-permutations of 𝑋?

𝑃 3,2 = 3 ⋅ 2 = 6

𝑎𝑏, 𝑎𝑐, 𝑏𝑎, 𝑏𝑐, 𝑐𝑎, 𝑐𝑏

Ex. How many way can we select a chairperson, vice-chairperson, secretary, and treasurer from a group of 10 person?

𝑃 10,4 = 10 ⋅ 9 ⋅ 8 ⋅ 7 = 5040

r-combination

Given a set 𝑋 = {𝑥1, 𝑥2, … , 𝑥𝑛} containing n distinct elements.

(a) an r-combination of 𝑋 is an unordered selection of r-elements of 𝑋(i.e., and r-element subset of 𝑋)

(b) The number of r-combination of a set of n distinct elements is denotes 𝐶(𝑛, 𝑟) or

𝑛𝑟

𝐶 𝑛, 𝑟 =𝑃 𝑛, 𝑟

𝑟!=

𝑛!

𝑛 − 𝑟 ! 𝑟!

Ex. A group of five students, Mary, Boris, Rosa, Ahmad, and Nguyen, has decided to talk with the Mathematics Department chairperson about having the Mathematics Department offer more courses in discrete mathematic. The chairperson has said that she will speak with three of the students. How many way can these five students choose three of their group to talk with chairperson?

𝐶 5,3 = 10

Ex. How many ways can we select a committee of three from a group of 10 distinct person?

𝐶 10,3 = 120

Ex. An ordinary deck of 52 cards consist of four suits: clubs, diamonds, hearts, spades, of 13 denominations each ace, 2-10, jack, queen, king.

(a)How many (unordered) five card poker hands, selected from an ordinary 52-card deck, are there?

𝐶 52,5 = 2,598,960

(b) How many poker hands contain cards all of the same suit?4 ⋅ 𝐶 13,5 = 5148

(c)How many poker hands contain three cards of one denomination and two cards of a second denomination?

13 ⋅ 12 ⋅ 𝐶 4,3 ⋅ 𝐶 4,2 = 3744

How many routes are there from the lower-left corner of an 𝑛 × 𝑛 square grid to the upper-right corner if we are restricted to travelling only to the right or upward?

𝐶(2𝑛, 𝑛)

Introduction to discrete probability

The probability 𝑃(𝐸) of an event 𝐸 for the finite sample space 𝑆 is

𝑃 𝐸 =𝐸

𝑆

Ex. Two fair dice are rolled. What is the probability that the sum of the numbers on the dice is 10?

Possible outcome for the sum is 10 : (4,6),(5,5),(6,4)

Outcome of the sample space S = 6×6

Thus

𝑃 𝐸 =3

36Ex. 5 microprocessors are randomly selected from a lot of 1000 microprocessors

among which 20 are defective. Find the probability of obtaining no defective microprocessors.

𝑆 = 𝐶 1000,5 , 𝐸 = 𝐶 980,5𝑃 𝐸 = 0.9037

Discrete Probability Theory

A probability function 𝑃 assigns to each out come 𝑥 in a sample space 𝑆 a number 𝑃(𝑥)

so that

0 ≤ 𝑃 𝑥 ≤ 1, for all 𝑥 ∈ 𝑆

and

𝑥∈𝑆

𝑃(𝑥) = 1

Ex. Suppose that a die is loaded so that the numbers 2 through 6 are equally likely to appear but that 1 is three times as likely as any other number to appear. To model this situation.

𝑃 2 = 𝑃 3 = 𝑃 4 = 𝑃 5 = 𝑃 6 , 𝑃 1 = 3𝑃 2

And

𝑃 2 =1

8, 𝑃 1 = 3

1

8=3

8

Probability of event

Let 𝐸 be an event. The probability of 𝐸, 𝑃(𝐸) is

𝑃 𝐸 =

𝑥∈𝐸

𝑃 𝑥

Ex. From the former example, find the probability of an odd number

𝑃 1 + 𝑃 3 + 𝑃 5 =3

8+1

8+1

8=5

8

Note :

For a fair die

𝑃 𝑜𝑑𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 =1

2

Probability of complement of event

Let 𝐸 be an event. The probability of ത𝐸, the complement of 𝐸 satisfies𝑃 𝐸 + 𝑃 ത𝐸 = 1

Union of event

Let 𝐸1 and 𝐸2 be events. Then𝑃 𝐸1 ∪ 𝐸2 = 𝑃 𝐸1 + 𝑃 𝐸2 − 𝑃(𝐸1 ∩ 𝐸2)

Corollary: union of mutual event

If 𝐸1 and 𝐸2 are mutually exclusive events𝑃 𝐸1 ∪ 𝐸2 = 𝑃 𝐸1 + 𝑃 𝐸2

Conditional probability

Let 𝐸 and 𝐹 be events, and assume that 𝑃 𝐹 > 0. The conditional probability of 𝐸 given 𝐹 is

𝑃 𝐸 𝐹 =𝑃 𝐸 ∩ 𝐹

𝑃 𝐹

Ex. E = die getting a sum of 10

F = at least one die show 5

𝐸 ∩ 𝐹: getting sum of 10 and at least one die show 5

𝑃 𝐸 ∩ 𝐹 =1

36

𝑃 𝐹 =11

36𝑃 𝐸 𝐹 = 1/11

Independent conditional event

Events E and F are independent if𝑃 𝐸 ∩ 𝐹 = 𝑃 𝐸 𝑃(𝐹)

Bay’s Theorem

Suppose that the possible classes are 𝐶1, … , 𝐶𝑛. Suppose further that each pair of classes is mutually exclusive and each item to be classified belongs to one of the classes. For a feature set 𝐹, we have

𝑃 𝐶𝑗 𝐹 =𝑃 𝐹 𝐶𝑗 𝑃 𝐶𝑗

σ𝑖=1𝑛 𝑃 𝐹 𝐶𝑖 𝑃 𝐶𝑖

Ex. At the telemarketing firm SellPhone , Dale, Rusty and Lee make calls. The following table shows the percentage of calls each caller makes and the percentage of the persons who are annoyed and hang up on each caller:

Let D denote the event “Dale made the call,” let R denote the event “Rusty made

the call.” let L denote the event “Lee made the call” and let H denote the event “the

caller hung up”. Find 𝑃(𝐷),

𝑃(𝑅), 𝑃(𝐿), 𝑃(𝐻|𝐷), 𝑃(𝐻|𝑅), 𝑃(𝐻|𝐿), 𝑃(𝐷|𝐻), 𝑃(𝑅|𝐻), 𝑃(𝐿|𝐻) and 𝑃(𝐻)

𝑃(𝐷) = 0.4, 𝑃(𝑅) = 0.25), 𝑃(𝐿) = 0.35𝑃(𝐻|𝐷) = 0.2, 𝑃(𝐻|𝑅) = 0.55, 𝑃(𝐻|𝐿) = 0.3

𝑃(𝐷|𝐻) use Bay’s Theorem

𝑃 𝐷 𝐻 =𝑃 𝐻 𝐷 𝑃 𝐷

𝑃 𝐻 𝐷 𝑃 𝐷 + 𝑃 𝐻 𝑅 𝑃 𝑅 + 𝑃 𝐻 𝐿 𝑃 𝐿= 0.248

Similar for 𝑃 𝑅 𝐻 = 0.426𝑃 𝐿 𝐻 = 1 − 𝑃 𝑅 𝐻 − 𝑃 𝐷 𝐻 = 0.326

Find 𝑃 𝐻 = 𝑃 𝐻 𝐷 𝑃 𝐷 + 𝑃 𝐻 𝑅 𝑃 𝑅 + 𝑃 𝐻 𝐿 𝑃 𝐿 = 0.3255

Caller

Dale Rusty Lee

Percent of calls 40 25 35

Percent of hang-ups 20 55 30

Generalized Permutations and Combinations

Theorem:

Suppose that a sequence 𝑆 of 𝑛 items has 𝑛1 identical objects of type 1, 𝑛2 identical objects of type 2,…, and 𝑛𝑡 identical objects type 𝑡. Then the number of orderings of 𝑆 is

𝑛!

𝑛1! 𝑛2! …𝑛𝑡!

Theorem:

If 𝑋 is a set containing 𝑡 elements, the number of unordered 𝑘-elements selections from 𝑋, repetitions allowed is

𝐶 𝑘 + 𝑡 − 1, 𝑡 − 1 = 𝐶(𝑘 + 𝑡 − 1, 𝑘)

Ex. How many strings can be formed using the following letters?𝑀 𝐼 𝑆 𝑆 𝐼 𝑆 𝑆 𝑃 𝑃 𝐼

Choose 2 P: C(11,2)

Choose 4 S: C(9,4)

Choose 4 I: C(5,4)

The outcome:𝐶 11,2 𝐶 9,4 𝐶 5,4 = 34,650

Binomial Coefficients and Combinatorial Identities

𝑎 + 𝑏 𝑛 = (𝑎 + 𝑏)(𝑎 + 𝑏)⋯(𝑎 + 𝑏)𝑛 𝑓𝑎𝑐𝑡𝑜𝑟𝑠

Ex.

𝑎 + 𝑏 3 = 𝑎 + 𝑏 𝑎 + 𝑏 𝑎 + 𝑏

= 𝑎𝑎𝑎 + 𝑎𝑎𝑏 + 𝑎𝑏𝑎 + 𝑎𝑏𝑏 + 𝑏𝑎𝑎 + 𝑏𝑎𝑏 + 𝑏𝑏𝑎 + 𝑏𝑏𝑏

= 𝑎3 + 𝑎2𝑏 + 𝑎2𝑏 + 𝑎𝑏2 + 𝑎2𝑏 + 𝑎𝑏2 + 𝑎𝑏2 + 𝑏3

= 𝑎3 + 3𝑎2𝑏 + 3𝑎𝑏2 + 𝑏3

𝑎 + 𝑏 𝑛 =𝑛

0𝑎𝑛𝑏0 +

𝑛1

𝑎𝑛−1𝑏1 +⋯+𝑛

𝑛 − 1𝑎1𝑏𝑛−1 +

𝑛𝑛

𝑎0𝑏𝑛

Binomial theoremTheorem :

𝑎 + 𝑏 𝑛 =

𝑘=0

𝑛𝑛𝑘

𝑎𝑛−𝑘𝑏𝑘

Ex. 𝑛 = 3

𝑎 + 𝑏 3 =30

𝑎3𝑏0 +31

𝑎2𝑏1 +32

𝑎1𝑏2 +33

𝑎0𝑏3

= 𝑎3 + 3𝑎2𝑏 + 3𝑎𝑏2 + 𝑏3

Ex. 3𝑥 − 2𝑦 4

=40

3𝑥 4 −2𝑦 0 +41

3𝑥 3 −2𝑦 1 +42

3𝑥 2 −2𝑦 2

+43

3𝑥 1 −2𝑦 3 +44

3𝑥 0 −2𝑦 4

= 34𝑥4 − 4 33𝑥32𝑦 + 6 32𝑥222𝑦2 − 4 3𝑥23𝑦3 + 24𝑦4

= 81𝑥4 − 216𝑥3𝑦 + 216𝑥2𝑦2 − 96𝑥𝑦3 + 16𝑦4

Ex. Find the coefficient of 𝑥2𝑦3𝑧4 in the expansion of 𝑥 + 𝑦 + 𝑧 9

We choose x 2 from 9 term, y 3 from 7 terms and z from 4 terms

Thus 92

73

44

=9!

2! 7!

7!

3! 4!=

9!

2! 3! 4!= 1260

Theorem𝑛 + 1𝑘

=𝑛

𝑘 − 1+

𝑛𝑘

, for 1 ≤ 𝑘 ≤ 𝑛

Proof

|𝑋| = 𝑛, choose 𝑎 ∉ 𝑋. Then 𝐶(𝑛 + 1, 𝑘) is the number of k-element of 𝑌 =𝑋 ∪ {𝑎}. The k-element of Y can be divided into 2 classes

1. Subset of Y not containing a

2. Subset of Y containing a

For class 1 are just k-element subset of 𝑋 =𝑛𝑘

For class 2 consist of (k-1) element subset of X together with a, there are 𝑛𝑘 − 1

.

Therefore 𝑛 + 1𝑘

=𝑛

𝑘 − 1+

𝑛𝑘

Ex. Show that

𝑖=𝑘

𝑛

𝐶(𝑖, 𝑘) = 𝐶 𝑛 + 1, 𝑘 + 1

Use Theorem in the form𝐶 𝑖, 𝑘 = 𝐶 𝑖 + 1, 𝑘 + 1 − 𝐶 𝑖, 𝑘 + 1

To obtain𝐶 𝑘, 𝑘 + 𝐶 𝑘 + 1, 𝑘 + 𝐶 𝑘 + 2, 𝑘 + ⋯+ 𝐶 𝑛, 𝑘

= 1 + 𝐶 𝑘 + 2, 𝑘 + 1 − 𝐶 𝑘 + 1, 𝑘 + 1 + 𝐶 𝑘 + 3, 𝑘 + 1− 𝐶 𝑘 + 2, 𝑘 + 1 +⋯+ 𝐶 𝑛 + 1, 𝑘 + 1 − 𝐶 𝑛, 𝑘 + 1= 𝐶(𝑛 + 1, 𝑘 + 1)

Ex. Find the sum

1 + 2 +⋯+ 𝑛

= 𝐶 1,1 + 𝐶 2,1 + ⋯+ 𝐶(𝑛, 1)= 𝐶 𝑛 + 1,2

=𝑛 + 1 𝑛

2

The Pigeonhole principle

• soon