Combinatorial Formula

Introduction:

Combinatorial is related to counting and arrangements of elements in sets.

Number of permutations of n elements: n!

Number of subsets of a set of n elements: 2n

Number of ordered arrangements of size r, from n elements, with replacement: nr, without replacement: P(n, r) = `(n!) / ((n - r)!)` .

Number of unordered arrangements of size r, without replacement, from a set of n elements:  C(n, r) = `(n!) / (r!(n-r)!)` . I like to share this Combinations Formula with you all through my article.

Example Problems:

Example 1: Ten groups participated in a dance competition, in how many ways dance competition can finish?

Solution:

To find the answer we have to find 10!.

10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 3628800

Therefore, the dance competition can finish in 3628800 ways.

Example 2: Set A = {a, b, c}, how many subsets does set A have?

Solution:

Set A has 3 elements, so number of subsets = 2n = 23 = 8

Subsets:

Subset with no element = {} = 1

Subset with 1 element = {a}, (b}, {c} = 3

Subset with 2 elements = {a, b}, {a, c}, {b, c} = 3

Subset with 3 elements = {a, b, c} = 1

1 + 3 + 3 + 1 = 8 = 23.

Example 3: In how many ways can 3 books be arranged from among 5 books with replacement?

Solution:

Number of ordered arrangements of size r, from n elements, with replacement: nr

Here n = 5, r = 3

nr = 53 = 125

Therefore, in 125 different ways 3 books can be arranged form among 5 books.

Example 4: In how many ways can 3 books be arranged from among 5 books without replacement?

Solution:

Number of ordered arrangements of size r, from n elements, without replacement:

P(n, r) = `(n!) / ((n - r)!)`

Here n = 5, r = 3

P(5, 3) = `(5!) / ((5 - 3)!)` = `(5!) / (2!)` = 60

Therefore, in 60 different ways 3 books can be arranged form among 5 books.

Example 5: In how many ways can 4 apples be chosen from among 6 apples without replacement?

Solution:

Number of unordered arrangements of size r, from n elements, without replacement:

C(n, r) = `(n!) / (r!(n - r)!)`

Here n = 6, r = 3

C(6, 3) = `(6!) / (3!(6 - 3)!)` = `(6!) / (3!xx3!)` = 20

Therefore, in 20 different ways 3 apples can be chosen form among 6 apples. Please express your views of this topic 6th grade math problems online by commenting on blog

Practice Problems:

Problem 1: Eight groups participated in a dance competition, in how many ways dance competition can finish?

Problem 2: Set A = {a, b, c, c, d}, how many subsets does set A have?

Problem 3: In how many ways can 4 books be arranged from among 7 books without replacement?

Problem 4: In how many ways can 5 apples be chosen from among 8 apples without replacement?

Answer: 1) 40320 2) 32 3) 840 4) 56

Formula Calculate Probability

Introduction to formula calculate probability:

Let us see about formula calculate probability. The probability is the division of the mathematics that agreement with manipulative the likelihood of a given event’s happening, which is articulated as a number among 1 and 0. The event with a probability of 1 can be measured a certainly.  The number of possible conclusion is named as probability. Understanding Calculate Percentile is always challenging for me but thanks to all math help websites to help me out.

Formula for Calculate Probability:

Let us see the formula for calculate the probability. The following formula is helped to calculate the probability.

By using Probability formula = `("Count of approving outcomes") / ("Total number of possible outcomes").`

Examples for Calculate Probability:

Let us see some examples of calculate the probability.

Example 1:

Consider 15 balls in a bucket. 5 of the 10 balls are blue. Calculate the probability to extract one blue ball from the bucket?

Solution:

There are 5 possible cases in a total of 10 balls.

By using Probability formula = `( "Count of approving outcomes") / ("Total number of possible outcomes")` .

The probability =` 5 / 10 ` = 0.5.

Thus the probability is calculated with the help of the formula.

Example 2:

One day 20 documents were presented to an income tax officer for disposal. Five documents contained bogus entries. All the documents were thoroughly mixed and there was no indication about bogus documents. What is the probability that one document with bogus entries is selected.

Solution:

Here all possible outcomes = 20

Let A be the event that the document has bogus entries.

Thus, number of possible outcomes = 5.

Here we will implement the classical definition of probability. All the 20 files are assumed to be regularly likely for the purpose of selecting a document. Please express your views of this topic online help with math problems by commenting on blog.

By using Probability of choosing a file with bogus entries is written as P(A).

P(A) = `( "Count of approving outcomes") / (" Total number of possible outcomes").`

P(A) = `5 / 10` = `1 / 4` .

Thus the probability is calculated with the help of the formula

Factoring Trigonometric Functions

Introduction to factoring trigonometric functions:

In mathematics, a rational function is any function which can be written as the ratio of two polynomial functions. Factorization (also factorization in British English) or factoring is the decomposition of an object (for example, a number, a polynomial, or a matrix) into a product of other objects, or factors, which when multiplied together give the original. (Source: Wikipedia)

Example Problems for Factoring Trigonometric Functions

Factoring trigonometric functions example problem 1:

Factoring the given trigonometric functions f (x) = cos^2x + 22cosx + 121

Solution:

Given equation is  f (x) = cos^2x + 22cosx + 121

Factorize the given equation, we get

f (x) = cos^2x + 11cosx + 11cosx + 121

Grouping the first two terms and second two terms, we get

f (x) = (cos^2x + 11cosx) + (11cosx + 121)

Take common terms in outside, we get

f (x) = cosx (cosx + 11) + 11 (cosx + 11)

= (cosx + 11) (cosx + 11)

The factors of the given trigonometric equation is (cosx + 11)2

Answer:

The final answer is (cosx + 11)2

Factoring trigonometric functions example problem 2:

Factoring the given trigonometric equations g (x) = sin^2x - 12sinx + 20

Solution:

Given equation is g (x) = sin^2x - 12sinx + 20

Factorize the given equation, we get

= sin^2x - 10sinx - 2sinx + 20

Grouping the first two terms and second two terms, we get

= (sin^2x - 10sinx) - (2sinx - 20)

Take common terms in outside, we get

= sinx (sinx - 10) - 2 (sinx - 10)

= (sinx - 10) (sinx - 2)

The factors of the given trigonometric equation is (sinx - 10) and (sinx - 2)

Answer:

The final answer is (sinx - 10) and (sinx - 2)

Understanding Derivative of Absolute Value Function is always challenging for me but thanks to all math help websites to help me out.

Practice Problems for Factoring Trigonometric Functions

Factoring trigonometric functions practice problem 1:

Factoring the given trigonometric functions f (x) = cos^2x - 7cosx + 6

Answer:

The final answer is (cosx - 6) and (cosx - 1)

Factoring trigonometric functions practice problem 2:

Factoring the given trigonometric functions g (x) = sin^2x + 12sinx - 28

Answer:

The final answer is (sinx + 14) and (sinx - 2)

Classifying Equations Math

Introduction to classifying equations math:

An equation is nothing but the mathematical statement, which asserts the equality of two expression. The equations forms the backbone of mathematics. Classifying the equations will helps to know the types of equations. In this article classifying equations math,  we are going to discuss about classifying the math equations.

Classifying Equations Math
Mathematics equations are classified into many types.  Those classifying equations are as follows:

Linear equations

Quadratic equations

Exponential equation

Cubic equation

Quartic equation

Constant equation

Explanation on Classifying the Equations:

Linear equations:

Linear equation is nothing but an algebraic equation. It has one or more variable. Each term of linear equation is either constant or multiplication of that constant.

Standard form of linear equation is y=ax +b

Quadratic equations:

Quadratic equation is second degree of expression or polynomial equation in algebra. The standard form of quadratic equation is . Please express your views of this topic how to graph linear equations by commenting on blog

ax2+bx+c=0

Here x is variable of the equations and a,b,c are constant of the equation must a?0.When a=0 it was called as linear.

Exponential equation:

Exponential function was can written as like ex, here e in constant number its approximate value is 2.7.This function was mainly used to change constant value in the independent variable, it gives same values of the  dependent  variable.

It can written as like y= ax^b

Cubic equation:

Cubic equation is third degree of expression or polynomial equation in algebra. The standard form of quadratic equation is

ax3+bx2+cx+d=0

Here x is variable of the equations and a, b, c ,d are constant of the equation.

Quartic equation:

Quartic equation is fourth degree of expression or polynomial equation in algebra. The standard form of quadratic equation is

Ax4+bx3+cx2+dx+e=0

Here x is variable of the equations and a, b, c ,d ,e are constant of the equation.

Constant equation:

Constant equation is nothing but having number alone. It can be written as X=8

Inconsistent System of Linear Equation

Introduction to inconsistent system of linear equation

In a system of linear equation, if there no one set of solution that can satisfy all the equations in the system then the system of linear equation are called inconsistent.

For example: 2x + y + 5 and 4x +2y = 7

Here if you try to solve these two equations, you will find that the variable get cancelled and become zero on the left side but then will be a consent on the right side. Here that is 0 = 3 which is not equal. Hence we cannot find a set of values to satisfy both the equations. Otherwise also, we can interpret this situation as they will be parallel lines, hence they won’t meet each other.

Now let us see few problems of this kind.

Example Problems on Inconsistent System of Linear Equation:

Ex 1: Check the following equations are consistent or inconsistent:

3x + 4y = 12 and 6x + 8y = 14.

Soln: Given lines are: 3x + 4y = 12  -------- 1

6x + 8y = 14 ----------2

(2)` xx ` (1)` implies`          6x + 8y = 24

(-)` xx` (2)` implies`             6x + 8y = 14` implies` 0 `!=` 10.

Since zero cannot be equal to 10, it is clear that the lines are inconsistent.

Ex 2: Check whether the following system of linear equations are consisted or inconsistent:     x + y = 5, x – y = 3.

Soln: Given:      x + y = 5    ---------------1

x – y = 3 -----------------2

Therefore (1 +2) `implies` 2x = 8 `implies` x = 4

Therefore       ` implies` 4 + y = 5  `implies` y =1

Since there is a unique value for x and y, (1) and (2) are consistent. Looking out for more help on Algebra Word Problems in algebra by visiting listed websites.

More Example Problems on Inconsistent System of Linear Equation:

Ex 3:  Check the consistency of the following equations:

` x/4` + `y/5 ` = 12 and `x/2` + `[2y]/5` = 24

Soln:     Given: `x/4` + `y/5` = 12   ----------------(1)

`x/2` + `(2y)/5` = 24   ----------------(2)

Therefore (1) `xx` (2) `implies` `2(x/4)` + `2(y/5)` = 2`xx` 12

`implies`` x/2` +`[2y]/5` = 24   ------(3)

Since (2) amd (3) are same, (1) and (2) represents the same line. Therefore they are consistent and have infinitely many solutions.

Polynomial in Standard Form

Introduction to polynomial in standard form:

A polynomial is either zero, or can be written as the sum of one or more non-zero terms The number of terms is finite. These terms consist of a constant  which may be multiplied by a finite number of variable. Each variable may have an  that is a non-negative integer, i.e., a natural numbers The exponent on a variable in a term is called the degreeof that variable in that term, the degree of the term is the sum of the degrees of the variables in that term, and the degree of a polynomial is the largest degree of any one term. Since x = x1, the degree of a variable without a written exponent is one. The degree of a constant term is 0. The coefficient of a term may be any number from a specified set.

Basic Concepts of Polynomial in Standard Form:

Terms of standard polynomial:

Standard form of polynomial is axn+bxn+cxn+d.It contains four terms and three variables and one constant.

Mononomial.

A polynomial contains only one terms is called as mononomial.

Example for monomial standard:

2x,8,3y2

Binomial.

A polynomial contains two terms mean it caaled as binomial.

Example for binomial standard form:

2x+6,3y+1,2x2+34

Trinomial

A polynomial contains three term mean called as trinomial

Example for trinomial standard form:

6x+7+4y,3x2+4x+9

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Polynomial in Standard Form-example Problem:

Example polynomial standard problems:

Polynomial Addition

Example 1: Find the sum of 4x4 – 7x2 + 8x + 4 and 5x + 7x3 – 2x2 +4

Solution:

(4x4 – 7x2 + 8x + 4) + (7x3 – 2x2 + 5x +4) = 4x4 + 7x3 – 7x2 – 2x2 + 8x + 5x + 4 +4

= 4x4 + 7x3 – (-9)x2 + (5+8)x + 4+4

= 4x4 + 7x3 +9x2 + 13x + 8

Polynomial subtraction

Example 2: Subtract 8x3 – 9x2 – 3 from 7x3 + 5x2 – 5x – 9.

Solution: (8x3 - 9x2 – 3) – (7x3 -5x+5x2 – 9)

8x3 - 9x2 – 3 – 7x3 +5x-5x2 + 9

= x3 -14x2 +5x – 3+9

=x3 -14x2 +5x +6

Answer x3 -14x2 +5x +6