Mathematics: 2016

Thursday, 11 August 2016

Topic 9: Sequence and Number pattern & Exercise

SEQUENCE AND NUMBER PATTERN

Introduction of sequence and number pattern
In mathematics, a sequence is an ordered list of objects. Like a set, it contains members (also called elements or terms). The number of ordered elements (possibly infinite) is called the length of the sequence. Unlike a set, order matters, and a particular term can appear multiple times at different positions in the sequence.

Understanding Sets

A set is a collection of objects, and it doesn't need to be a number!
This is the set of the clothes in my closet: C = {pants, t-shirt, skirt, and dress}. The capital C represents the set. So, if I said set C, we know I'm talking about clothes in my closet. The braces, { }, denote the elements, or members of the set. The elements of set C are pants, t-shirt, skirt, and dress.

You're probably familiar with a set of real numbers: R = {…-3, -2, -1, 0, 1, 2, 3...}. The three dots indicate that the pattern continues. The elements of this group are all real numbers. So, R equals the set of real numbers.

A mathematical sequence is an ordered list of objects, often numbers. Sometimes the numbers in a sequence are defined in terms of a previous number in the list.

Source: Boundless. “Introduction to Sequences.” Boundless Algebra Boundless, 21 Dec. 2016. Retrieved 22 Jan. 2017 from https://www.boundless.com/algebra/textbooks/boundless-algebra-textbook/sequences-and-series-342/sequences-and-series-53/introduction-to-sequences-224-5904/

It is easiest to start by showing the growth of a simple repeating pattern.

Simple repeating pattern

Show how it grows by adding successive identical units of repeat.

Growing pattern

Counting the number of blocks give the sequence 2, 4, 6...people can see that each time a unit of repeat is added, the total number of blocks increase by 2. Also the total number is twice the number of unit of repeat.

Example 1:

1, 4, 7, 10, 13, 16......start at 1 and jumps 3

Example 2:

1, 4, 7, 10, 13, 16, 19, 22, 25, ...

This sequence has a difference of 3 between each number.
The pattern is continued by adding 3 to the last number each time, like this:

Example 3:

3, 8, 13, 18, 23, 28, 33, 38, ...

This sequence has a difference of 5 between each number.

The pattern is continued by adding 5 to the last number each time, like this:

This video is to practice finding pattern in number.

Topic 8: Linear Programming & Exercise

LINEAR PROGRAMMING

Introduction of Linear Programming
Linear programming is the process of taking various linear inequalities relating to some situation and finding the best value obtainable under those conditions. A typical example would be taking the limitations of materials and labor and then determining the best production level for maximal profits under those conditions.

In real life, linear programming is part of a very important area of mathematics calles "optimization techniques". This field of study are used every day in the organization and allocation of resource. These real life systems can have dozens or hundreds of variable or more. in algebra through you will only work with the simple (and graph able) two variable linear case.

The general process for solving linear-programming exercises is to graph the inequalities (called the "constraints") to form a walled-off area on the x,y-plane (called the "feasibility region"). Then you figure out the coordinates of the corners of this feasibility region (that is, you find the intersection points of the various pairs of lines), and test these corner points in the formula (called the "optimization equation") for which you're trying to find the highest or lowest value.

Formula of linear Programming

Y > Mx + C

M = gradient

C = y-intercept

Example 1:

A company makes two products (X and Y) using two machines (A and B). Each unit of X that is produced requires 50 minutes processing time on machine A and 30 minutes processing time on machine B. Each unit of Y that is produced requires 24 minutes processing time on machine A and 33 minutes processing time on machine B.
At the start of the current week there are 30 units of X and 90 units of Y in stock. Available processing time on machine A is forecast to be 40 hours and on machine B is forecast to be 35 hours.
The demand for X in the current week is forecast to be 75 units and for Y is forecast to be 95 units. Company policy is to maximize the combined sum of the units of X and the units of Y in stock at the end of the week.

Formulate the problem of deciding how much of each product to make in the current week as a linear program.
Solve this linear program graphically.

Solution

Let

x be the number of units of X produced in the current week
y be the number of units of Y produced in the current week

then the constraints are:

50x + 24y <= 40(60) machine A time

The objective is: maximize (x+30-75) + (y+90-95) = (x+y-50)
i.e. to maximize the number of units left in stock at the end of the week
It is plain from the diagram below that the maximum occurs at the intersection of x=45 and 50x + 24y = 2400

30x + 33y <= 35(60) machine B time

x >= 75 - 30

i.e. x >= 45 so production of X >= demand (75) - initial stock (30), which ensures we meet demand

y >= 95 - 90

i.e. y >= 5 so production of Y >= demand (95) - initial stock (90), which ensures we meet demand.

Example 2:

A company manufactures two products (A and B) and the profit per unit sold is £3 and £5 respectively. Each product has to be assembled on a particular machine, each unit of product A taking 12 minutes of assembly time and each unit of product B 25 minutes of assembly time. The company estimates that the machine used for assembly has an effective working week of only 30 hours (due to maintenance/breakdown).
Technological constraints mean that for every five units of product A produced at least two units of product B must be produced.

Formulate the problem of how much of each product to produce as a linear program.
Solve this linear program graphically.
The company has been offered the chance to hire an extra machine, thereby doubling the effective assembly time available. What is the maximum amount you would be prepared to pay (per week) for the hire of this machine and why?

Solution

Let
x_A = number of units of A produced
x_B = number of units of B produced
then the constraints are:
12x_A + 25x_B <= 30(60) (assembly time)
x_B >= 2(x_A/5)
i.e. x_B - 0.4x_A >= 0
i.e. 5x_B >= 2x_A (technological)
where x_A, x_B >= 0
and the objective is
maximize 3x_A + 5x_B
It is plain from the diagram below that the maximum occurs at the intersection of 12x_A + 25x_B = 1800 and x_B - 0.4x_A = 0

Example 3:

If the objective function of the previous exercise had been:

f(x,y) = 20x + 30y

f(0,500) = 20·0 + 30 · 500 = $15,000 Maximum

f(500, 0) = 20·500 + 30·0 = $10,000

f(375, 250) = 20·375 + 30·250 = $15,000 Maximum

In this case, all the pairs with integer solutions of the segment drawn in black would be the maximum.

Below is the video to show you how to solve the linear programming with the graph.

Topic 5: Permutation and Combination & Exercise

...PERMUTATION...

What is permutation?? People always get confused "permutation" and "combination" which one's which??

Introduction of permutation

Permutation sound complicated, doesn't?? and it is "Permutation is every little detail matters". Permutation actually an arrangement of all or part of a set of objects, with regard to the order of the arrangement.

In mathematics, the notion of permutation relates to the act of arranging all the members of a set into some sequence or order or if the set is already ordered, rearranging (reordering) its elements, a process called permuting. These differ from combinations, which are selections of some members of a set where order is disregarded.

Meaning of permutation

A permutation is an arrangement, or listing, of objects in which the order is important. In the previous lessons, we looked at examples of the number of permutations of n things taken n at a time. Permutation is used when we are counting without replacement and the order matters. If the order does not matter then we can use combinations.

Permutations occur, in more or less prominent ways, in almost every area of mathematics. They often arise when different orderings on certain finite sets are considered, possibly only because one wants to ignore such orderings and needs to know how many configurations are thus identified. For similar reasons permutations arise in the study of sorting algorithms in computer science.

$_{n}^{r} P = \frac{n!}{(n - r)!}$ What is the Permutation Formula?
In general P(n, r) means that the number of permutations of n things taken r at a time. We can either use reasoning to solve these types of permutation problems or we can use the permutation formula.

The formula for permutation is

Example 1:

A zip code contains 5 digits. How many different zip code can be made with the digit 0 - 9 if no digit is used more than once and the first digit is not 0?

Solution:

Using reasoning:

for the first position, there are 9 possible choice (since 0 is not allowed). After that number is chosen, there are 9 possible choices (since 0 is not allowed). Then there are 8 possible choices, 7 possible choices and 6 possible choices.

9 x 9 x 8 x 7 x 6 = 27,216

Using permutation formula:

We can't include the first digit in the formula because 0 is not allowed. For the first position, there are 9 possible choices (since 0 is not allowed). For the next 4 possitions, we are selecting from 9 digits.
9 x P(9,4) = 9 x 9! / (9 - 4)! = 9 x 9! / 5! = 9 x 9 x 8 x 7 x 6 x5! / 5! = 27,216

Example 2:

if five digit 1,2, 3, 4, 5 are being given and a three digit code has to be made from it if the repetition of digits is allowed then how many such code can be formed.

Solution:

As repetition is allowed, we have five options for each digit of the code. Hence the required number of ways code can be formed is 5 x 5 x 5 = 125.

Example 3:

if three alphabets are to be chosen from A, B, C, D and E such that repetition is not allowed then in how many ways it can be dome?

Solution:

The number of ways three alphabets can be chosen from five will be,

3 / 5p = 5! / (5-3)! = 5 x 4 x 3 x 2 x1 / 2 x 1 = 60

Hence, there are 60 possible ways.

Permutations Word Problems

Example 1:

In how many ways can the letters of the word APPLE can be rearranged?

Solution:

Total number of alphabets in APPLE = 5.

Number of repeated alphabets = 2

Number of ways APPLE can be rearranged =

\frac{5!}{2!}

= 60.

The word APPLE can be rearranged in 60 ways.

Example 2:

10 students have appeared in a test in which the top three will get a prize. How many possible ways are there to get the prize winners?

Solution:

We need to choose and arrange 3 persons out of 10. Hence, the number of possible ways will be

_{10}^{3} P

\frac{10!}{(10 - 3)!}

10 \times 9 \times 8

720

Example 3:

Ellie want to change her password which is ELLIE9 but with same letters and number. In how many ways she can do that?

Solution:

Total number of letters = 6.

Repeated letters = 2 Ls and 2 Es.

Number of times ELLIE9 can be rearranged =

\frac{6!}{2! 2!}

6 \times 5 \times 3 \times 2 \times 1

180

.

But the password need to be changed. So, the number of ways new password can be made =

180 - 1 = 179

This is the video for permutation. let's learn from the basic..

...COMBINATION...

Introduction:

Each of the different groups or selections which can be formed by taking some or all of a number of objects is called a combination. Combination doesn't have any rules.

a combination is a way of selecting items from a collection, such that (unlike permutation) the order of selection does not matter. In smaller cases it is possible to count the number of combinations. For example, given three fruits, say an apple, an orange and a pear, there are three combinations of two that can be drawn from this set: an apple and a pear; an apple and an orange; or a pear and an orange. More formally, a k-combination of a set S is a subset of k distinct elements of S.

$\textstyle {\frac {n!}{k!(n-k)!}}$

FORMULA FOR COMBINATION

$C (n, k)$ = $\frac{n!}{(k! (n - k)!)}$

Example 1:

In a lucky draw chits of ten names are out in a box out of which three are to be taken out. Find the number of ways in which those three names can be taken out.

Solution: The possible number of ways for finding three names out of ten from the box is:

C (10, 3) =

\frac{10!}{(3! 7!)}

\frac{10 * 9 * 8 * 7!}{7! * 3 * 2 * 1}

\frac{720}{6}

= 120

So there are 120 different ways of choosing three names out of the ten from the box.

Example 2:

Let us suppose we have 12 adults and 10 kids as an audience of a certain show. Find the number of ways the host can select three persons from the audiences to volunteer. The choice must contain two kids and one adult.

Solution: As order here does not matter so we have:

C (10, 2) * C (12, 1) = [10 *

\frac{9}{2}

] * [

\frac{12}{1}

] = 45 * 12 = 540.

So there are 540 ways in which the host can choose the volunteers containing two kids and an adult.

Example 3:

Let us suppose we have 12 adults and 10 kids as an audience of a certain show. Find the number of ways the host can select three persons from the audiences to volunteer. The choice must contain two kids and one adult.

Solution: As order here does not matter so we have:

C (10, 2) * C (12, 1) = [10 *

\frac{9}{2}

] * [

\frac{12}{1}

] = 45 * 12 = 540.

So there are 540 ways in which the host can choose the volunteers containing two kids and an adult.

Below is the video for combination. 👇

Exercise

1) what is 8!

2)What is 5P5? 3) A word is chosen at random from the word FEBRUARY. Find the probability that the word chosen at random is, a) Find the letter E?

Mathematics

Thursday, 11 August 2016

Topic 9: Sequence and Number pattern & Exercise

SEQUENCE AND NUMBER PATTERN

Understanding Sets

Tuesday, 9 August 2016

Topic 8: Linear Programming & Exercise

LINEAR PROGRAMMING

Solution

Solution

Monday, 11 July 2016

Topic 5: Permutation and Combination & Exercise

...PERMUTATION...