Different Number Combinations

In combinatorial mathematics, a k-combination of a finite set S is a subset of k different numbers of S. Specifying a subset does not arrange them in a particular order; by contrast, producing the k different numbers in a specific order defines a sequence without repetition, also called k-permutation (but which is not a permutation of S in the usual sense of that term. SOURCE: WIKIPEDIA



Example problems of different number combinations:

Different number combinations example 1:

How many lines can you draw using THREE non collinear (not in a single line) points X, Y and Z on a plane?

Solution:

You need two points to draw a line. The order is not important. Line XY is the same as line YX. The problem is to select TWO points out of THREE to draw different lines. If we proceed as we did with permutations, we get the following pairs of points to draw lines.

XY, XZ

YX, YZ

ZX, ZY

There is a problem: line XY is the same as line YX, same for lines XZ and ZX and YZ and ZY.

The lines are: XY, YZ and XZ; three lines only.

So in fact we can draw THREE lines and not SIX and that's because in this problem the order of the points X, Y and Z is not important.

This is a combination problem: combining TWO items out of THREE and is written the general form as follows:

n C r = n! / [ (n - r)! r! ]

Special case:

n C 0 = n C n = 1

The number of combinations (nCr) is equal to the number of permutations divided by r! to removes those counted more than once because the order is not mainly use.

Different number combinations example 2:

Calculate 4C3

Solution:

We can find the combination value by using the following formula:

n C r = n! / [(n - r)! r!]

Substitute the value of n and r into the above formula, then we get

4C3= 4! / [(4-3)! 3!]

=4! / [1!*3!]

=24/[1*6]

=24/6

=4

Answer: 4

Different number combinations example 3:

Calculate 6C6

Solution:

We can find the combination value by using the following formula:

n C r = n! / [(n - r)! r!]

Substitute the value of n and r into the above formula, then we get

6C6= 6! / [(6-6)! 6!]

=6! / [0!*6!]

=6! / [1*(6!)]

=6! / 6!

= 1

Answer: 1

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Practice problems of different number combinations:


1) In how many ways can you select a committee of 2 students out of 5 students?

Answer: 20 (5C2)

2)  How many triangles can you make using 4 non collinear points on a plane?

Answer: 24 (4C3)

Learn Area of a Rhombus

Rhombus is a quadrilateral in which all the sides are equal. Rhombus has all the properties of a parallelogram and its diagonals bisect each other at right angles. The area of a rhombus can be calculated using the conventional formula used to calculate the area of a parallelogram b*h where "b" is the base and "h" is the perpendicular distance between the base and the opposite parallel line.



ABCD is a rhombus and we shall derive the formula for finding out the area given its diagonals AC and BD


learning how to find area of a rhombus


Let the length AC be d1 and that of BD be d2. The formula for calculating the area of a triangle is 1/2 bh where 'b" is the base and "h" is height or altitude.

We shall consider the rhombus as the combinations of two triangles ABC and ADC.If we add the areas of triangles ABC and ADC we are sure to get the area of the rhombus ABCD.

The diagonals of rhombus bisect each other at right angles the height of triangles of ABC and ADC will be half of the diagonal BD which is equal to d2/2. We shall now calculate the area separately and add.


Formula to learn area of a rhombus


Area of triangle ABC =1/2  (d1* d2/ 2) =(d1 d2) /4

Area of triangle ADC=1/2  (d1* d2/ 2) =(d1 d2) /4

Area ABC +Area ADC = (d1 d2 )/4 + (d1 d2 ) /4 =d1 d2 /2

The area of the rhombus  = 1/2 product of its diagonals

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learn examples on area of a rhombus


We shall illustrate this by two examples. We are given the diagonals of a rhombus as 10 cms and 12 cms. We are asked to find the area of the rhombus. The area of a rhombus= 1/2 the product of its diagonals= 1/2 *10*12 = 60 square cm.

This is an example to find the area of the rhombus using the conventional formula. Suppose the length of the side of a rhombus is given as 12cm and the perpendicular distance between the sides is given as 10 cm the area of the rhombus =side* perpendicular distance = 10 * 12 =120square cm.

Rational Algebraic Expressions

A rational expression is an algebraic expression of the `(P)/(Q)` from ,where P and Q are simpler expressions (usually polynomials), and the denominator Q is not zero.



This rational expressions whose numerators and denominators are (or written as) polynomials. Like polynomials, rational expressions appear frequently in the Algebra II and higher mathematics. We must understand the how to perform basic operations with rational expressions and how to solve rational equations.


Two Types:-


1.Adding and Subtracting Rational Expressions

2.Multiplying and Dividing Rational Expressions



1. Adding and Subtracting Rational Expressions

Adding Subtracting Rational Expressions with the Like Denominators

Add and subtract their numerators and then write the results over the denominator. Then, simplify the numerator factor, and write the expression in lowest terms.

Adding and Subtracting Rational Expressions with the Unlike Denominators

Rational expressions can be added and subtracted if and only if they have the same denominator. Thus, we must rewrite them as expressions with a common denominator.



2.Multiplying and Dividing Rational Expressions

Multiplying Rational Expressions

To multiply two rational expressions, factor them. Then multiply their numerators and denominators, crossing out the any factors that appear in both the numerator and the denominator

Dividing Rational Expressions

To divide by rational expression,and multiply by its reciprocal.


Examples:-


Example 1:

Simplify the rational expression

4x - 2
--------
2x - 1
Solution :-

Factor both the numerator and denominator completely.

2(2x - 1)
---------
2x - 1
Cancel common factors to reduce and then simplify the given expression.

2(2x - 1)
= ------------ = 2 , with x not equal to 1
2x - 1


Example 2:

Simplify the rational expression

4x + 16
---------
x2 - 16
Solution:-

Factor both the numerator and denominator completely.

4x + 16         4(x + 4)
--------- = ----------------
x2  - 16      (x + 4)(x - 4)
Cancel common factors to simplify the expression.

4(x + 4)
= ---------------
(x - 4)(x + 4)


4
= --------    , with x not equal to -4.
(x - 4)

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Example 3:

Simplify the rational expression

x2 + x - 2
-------------
-x2 -2x + 3

Solution:-

Factor both the numerator and denominator completely.

x2 + x - 2      (x + 2)(x - 1)
------------- = ----------------
-x2 -2x + 3      (x + 3)(-x + 1)


Note that - x + 1 = - (x - 1) in the denominator.

(x + 2)(x - 1)
= -----------------
-(x + 3)(x - 1)


Cancel common factors.

(x + 2)(x - 1)
=--------------------
-(x + 3)(x - 1)


(x + 2)
= -------------   , with x not equal to 1
(x + 3)

Learning Compound Interest

Compound interest is paid on the principal and also for the interest accumulated in the past years. Compound interest emerges when interest is added to the principal, so the interest that has been added also earns interest. This adding of interest with the principal is said to be compounding. Compounding can be done for a time like yearly, quarterly, monthly, daily etc.


Learning Compound Interest Formula:


Learning The basic formula for Compound Interest is:

FV = PV (1+r)n

PV is the present value

r is the annual rate of interest (percentage)

n is the number of years the amount is deposit or borrowed for.

FV = Future Value is the amount of money accumulate after n years, including interest.

Learning Frequent Compounding of Interest:

If interest is paid more frequently, Here are a few examples of the formula:

Annually = P × (1 + r) = (annual compounding)

Quarterly = P (1 + r/4)4 = (quarterly compounding)

Monthly = P (1 + r/12)12 = (monthly compounding)

Learning Find the Present Value when know a Future Value, the Interest Rate and number of Periods.

PV = FV / (1+r)n

Learning Find the Interest Rate when know the Present Value, Future Value and number of Periods.

r = ( FV / PV )1/n – 1

Learning Find the number of Periods when know the Present Value, Future Value and Interest Rate

n = ln(FV / PV) / ln(1 + r)

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Example of compound interest


Ex 1: I have $1000.00 to invest for 3 years at rate of 6% compound interest.

Sol:  Here p=$1000, n=3, r=5

A = 1000 (1 + 0.06)3 = $1191.02.

1000.00 is worth $1191.02

Ex 2:  How many years to turns 1000into10,000 at 5% interest?

Sol :    n = ln( 10000 / 1,000 ) / ln( 1 + 0.05 ) = ln(10)/ln(1.05) = 2.3026/0.04879 = 47.19

Ex 3 :  I have $2000.00 to invest for 3 years at rate of 7% compound interest.

Here p=$2000, n=3, r=7

A = 2000 (1 + 0.07)3 = $2450.09.

2000.00 is worth $2450.09.

Angle of Elevation Learning

Introduction:

Assume that you are walking on a road. Normally your eyes will be viewing objects at a horizontal level at a distance of your height from the ground. Suppose you see a tall building in the vicinity and you want to see the top of the building. What you do is, rotate your line of sight from horizon to spot the top of thr building. The angle by which you do that is called the Angle of Elevation.

The angle of elevation and depression are a wonderful concepts in the subject of trigonometry.


Angle of elevation and depression- Definition


Let AB be an object. The angle θ that is required to see the point A from point O is called the angle of elevation of the object at point O.

The angle of elevation of an object depends on the point from where it is measured. The more you are closer to the object, the more is the angle of elevation and vice versa.

Hence, it is a must that the angle of elevation of an object should always be accompanied by the information from where it is measured.

Angle of Depression is the angle by which you tilt your eye down from horizontal line to see an object which is below you.


Angle of elevation – Practical applications

The concept of angle of elevation and depression is used very widely in estimations of heights and depths. In fact, in trigonometry, a separate topic heights and distances deals with angle of elevation and angle of depression. The angle of depression is the angle made below the horizon for objects at lower levels.



Suppose the height of the building shown in the above picture (h) is to be estimated. It may not be practicable to measure that directly. But it is always possible to measure the angle of elevation θ (equipment for such equipment is available) and also the horizontal distance between the building and the point of angle measurement (d).

The basic trigonometric identity, tan θ =`(h)/(d)`, will easily give you the estimate.


Problems based on Angle of elevation and angle of depression


Let us do a  problems relating to angle of elevation and angle of depression to understand the concept properly.

Angle of Elevation Problems :

1)  A balloon is flying high. A boy looks at it. His eyes make an angle of 30 º from the ground to the balloon.

This is angle of elevation. The distance between the boy and  the flying balloon is 10 m. Find the height of the balloon above the Ground.

Here we use sine 30º formula =  opposite side / hypotenuse   = height / hypotenuse

Sine 30º= 1/2     hypotenuse = 10 m

so height = sine 30º x 10  =    `(1)/(2)` x 10m =  5 meter

So the ballon flies at a height of 5 meter above the ground

Problems based on Angle of depression

Problem 2)From the top of a light house, the angle of depression of 2 ships on the opposite side of the light house  are

observed as  45º and 3 angle 0º. The height of the light house is 200 m. Find the distance between the ships.


Let  M be  a ship and N be another ship.   Let AB be the light house

CD is a line drawn horizontally through B.

From B  the angle of depression is 45º and 30º.

CD is parallel to MN

So angle CBM= angle BMA and angle DBN = angle BNA.

That makes angle BMA=45º and angle BNA= 30 degrees(Interior alternate angles  are equal)

In the right triangle BAN ,  tan 30º = AB/AN

Tan 30º= `(1)/(sqrt(3))`        So `(1)/(sqrt(3))`  = `(200)/(AN)`

So AN =  200√3 = 200 x 1.732 = 346.4 m

Now let us measure AM

Tan 45º = 1   Tan 45º = `(AB)/(AM)`  so AM x 1 = AB   AM= AB   AB= 200m so AM = 200 m

Distance between Ship M and the light house is 200m

Distance between ship N and the light house is 346.4 m

Add them both.

Distance between ship M and N is 200+346.4 = 546.4 meters

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Probability Problems with Dice Eighth Grade

Probability is the chance that something will happen - how likely it is that some event will happen.If a trial will produce N commonly exclusive and equally probable outcomes out of which n outcomes are positive to the amount of event A, then the probability of A is mentioned by P(A) and is defined as the ratio n/ N. Thus the probability of A is given by

                                           Event (P)   = Number of Successful outcomes
                                                                   Number of total outcomes

probability problem with dice for 8th grade:

Problem:

A fair die is thrown. Find the probabilities that the face on the die Is 

 (1) Maximum (2) Prime (3) Multiple of 3 (4) Multiple of 7

Solution:
            In the above probability problem there are 6 possible outcomes when a die is tossed. We assumed that all the 6 faces are equally likely. The classical definition of probability is to be applied here
            

The sample space is ,S={1,2,3,4,5,6}  => n(S)=6

(1) Let A be the event that the face is maximum
            Thus,
                       A=(6),        n(A)=1
            Therefore, P(A) = n(A) / n(S)

                                  = 1/6

(2) Let B be the event that the face is prime

            Thus,
                       B=(2,3,5)        n(B)=3   
            Therefore, P(B) = n(B) / n(S)

                                  = 3/6

                                  => 1/2

(3) Let C be the event that the face is multiple of 3
            Thus,          C=(3,6)    n(C)=2
          Therefore, P(C) = n(C) / n(S)

                                = 2/6

                                =>1/3

(4) Let D be the event that the face is multiple of 7
            Thus,          D  =0,           n(D)=0
           Therefore,P(D)  = n(D)/n(S)

                                  =0/6

                                  =0   (not possible)

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Dice problem for 8th grade:

Example:

Solve the problem for probability that when 2 standard 6-sided dice are rolled, the sum of the numbers on the top faces is 4.

Solution:

In the above problem, A dice contains 6 faces, so the possible out comes from the  both dices are      

                            = 6 * 6 
                            = 36 total number of possible outcomes 
Here the given condition is “Some of the numbers on the top of the face is 4”.
So the possible out comes are 
                   Dice 1             Dice 2
                      1         +         3           = 4
                      2         +         2           = 4
                      3         +         1           = 4
the total number of successful outcomes are 3

Final solution 
                          P (sum of 4)     = 3 / 36

                                                 = 1 / 12.

Volume of a Cylinder

Volume of a cylinder is a measurement of the occupied units of a cylinder. The volume of a cylinder is represented by cubic units like cubic centimeter, cubic millimeter and so on. Volume of a cylinder is the number of units used to fill a cube.

A cylinder has two parallel faces in its structure. The height of a cylinder is perpendicular to the two bases. The cylinder has a round base and a perpendicular height.



The volume of a cylinder formula depends on the area of the cylinder. The formula for volume of a cylinder can be written as,

Volume of cylinder = pi * r2 * h.


Formula for Surface Area of a Cylinder


The formula for volume of a cylinder depends on the area of the cylinder. The area of a cylinder can be written as, A = pi * r2. Here, pi is the constant value and r is the radius of the cylinder. To find the volume of the cylinder, we have to multiply the height of the cylinder to the area of the cylinder. The volume of cylinder formula can be written as,

Volume of cylinder = pi * r2 * h.

Here, pi is the constant variable, r is the radius of the cylinder and h is the height of the cylinder. The constant variable pi is equal to (22 / 7) or 3.14.


How to Find The Volume of a Cylinder


Given below are some of the examples to find the volume of a cylinder.

Example 1:

Calculate volume of a cylinder having the radius of cylinder is 3cm and height is 4cm.

Solution:

The given radius of the cylinder is 3cm and height of the cylinder is 4cm

Formula:

Volume of a cylinder = pi * r2 * Height

Volume of cylinder = 3.14 * 32 * 4 cm3

= 113.04 cm3

Answer:

The volume of cylinder is 113. 04 cm3

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Example 2:

The radius and height of the cylinder is 5 cm, 21 cm. Find the volume of cylinder.

Solution:

Given radius r = 5 cm, height = 21 cm.

Formula:

Volume of cylinder = pi * r2 * h

= 3.14 * 52 * 21 cm3

= 1648.5 cm3

Answer:

The volume of given cylinder is 1648.5 cm3

Example 3:

Find the volume of cylinder for the given base and height. base = 6 cm and height = 4 cm

Solution:

Given base = 6 cm and height = 4 cm

Formula:

Volume of cylinder = pi * r2 * h

= 3. 14 * 62 * 4 cm3

= 452.16 cm3

Answer:

The volume of cylinder is 452.16 cm3

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