Permutation and Combination

Permutation : Permutation means arrangement of things. The word arrangement is used, if the order of things is considered.

Combination: Combination means selection of things. The word selection is used, when the order of things has no importance.

Example: Suppose we have to form a number of consisting of three digits using the digits 1,2,3,4, To form this number the digits have to be arranged. Different numbers will get formed depending upon the order in which we arrange the digits. This is an example of Permutation.

Now suppose that we have to make a team of 11 players out of 20 players, This is an example of combination, because the order of players in the team will not result in a change in the team. No matter in which order we list out the players the team will remain the same! For a different team to be formed at least one player will have to be changed.

Now let us look at two fundamental principles of counting:

Addition rule : If an experiment can be performed in ‘n’ ways, & another experiment can be performed in ‘m’ ways then either of the two experiments can be performed in (m+n) ways. This rule can be extended to any finite number of experiments.

Example: Suppose there are 3 doors in a room, 2 on one side and 1 on other side. A man want to go out from the room. Obviously he has ‘3’ options for it. He can come out by door ‘A’ or door ‘B’ or door ’C’.

Multiplication Rule : If a work can be done in m ways, another work can be done in ‘n’ ways, then both of the operations can be performed in m x n ways. It can be extended to any finite number of operations.

Example.: Suppose a man wants to cross-out a room, which has 2 doors on one side and 1 door on other site. He has 2 x 1 = 2 ways for it.

Factorial n : The product of first ‘n’ natural numbers is denoted by n!.

n! = n(n-1) (n-2) ………………..3.2.1.

Ex. 5! = 5 x 4 x 3 x 2 x 1 =120

Note 0! = 1

Proof n! =n, (n-1)!

Or (n-1)! = [n x (n-1)!]/n = n! /n

Putting n = 1, we have

O! = 1!/1

or 0 = 1

Permutation

Number of permutations of ‘n’ different things taken ‘r’ at a time is given by:-

ⁿP_r = n!/(n-r)!

Proof: Say we have ‘n’ different things a₁, a₂……, a_n.

Clearly the first place can be filled up in ‘n’ ways. Number of things left after filling-up the first place = n-1

So the second-place can be filled-up in (n-1) ways. Now number of things left after filling-up the first and second places = n - 2

Now the third place can be filled-up in (n-2) ways.

Thus number of ways of filling-up first-place = n

Number of ways of filling-up second-place = n-1

Number of ways of filling-up third-place = n-2

Number of ways of filling-up r-th place = n – (r-1) = n-r+1

By multiplication – rule of counting, total no. of ways of filling up, first, second -- rth-place together :-

n (n-1) (n-2) ------------ (n-r+1)

Hence:
ⁿP_r = n (n-1)(n-2) --------------(n-r+1)

= [n(n-1)(n-2)----------(n-r+1)] [(n-r)(n-r-1)-----3.2.1.] / [(n-r)(n-r-1)] ----3.2.1

ⁿP_r = n!/(n-r)!

Number of permutations of ‘n’ different things taken all at a time is given by:-

ⁿP_n = n!

Proof :
Now we have ‘n’ objects, and n-places.

Number of ways of filling-up first-place = n

Number of ways of filling-up second-place = n-1

Number of ways of filling-up third-place = n-2

Number of ways of filling-up r-th place, i.e. last place =1

Number of ways of filling-up first, second, --- n th place
= n (n-1) (n-2) ------ 2.1.

ⁿP_n = n!

Concept.

We have ⁿP_r = n!/n-r

Putting r = n, we have :-

ⁿP_r = n! / (n-r)

But ⁿP_n= n!

Clearly it is possible, only when n! = 1

Hence it is proof that 0! = 1

Note : Factorial of negative-number is not defined. The expression –3! has no meaning.

Examples

Q. How many different signals can be made by 5 flags from 8-flags of different colours?

Ans. Number of ways taking 5 flags out of 8-flage = ⁸P₅

= 8!/(8-5)!

= 8 x 7 x 6 x 5 x 4 = 6720

Q. How many words can be made by using the letters of the word “SIMPLETON” taken all at a time?

Ans. There are ‘9’ different letters of the word “SIMPLETON”

Number of Permutations taking all the letters at a time = ⁹P₉

= 9! = 362880.

Number of permutations of n-thing, taken all at a time, in which ‘P’ are of one type, ‘g’ of them are of second-type, ‘r’ of them are of third-type, and rest are all different is given by :-

n!/p! x q! x r!

Example: In how many ways can the letters of the word “Pre-University” be arranged?

13!/2! X 2! X 2!

Number of permutations of n-things, taken ‘r’ at a time when each thing can be repeated r-times is given by = n^r.

Proof.

Number of ways of filling-up first –place = n

Since repetition is allowed, so

Number of ways of filling-up second-place = n

Number of ways of filling-up third-place

Number of ways of filling-up r-th place = n

Hence total number of ways in which first, second ----r th, places can be filled-up

= n x n x n ------------- r factors.

= n^r

Example: A child has 3 pocket and 4 coins. In how many ways can he put the coins in his pocket.

Ans. First coin can be put in 3 ways, similarly second, third and forth coins also can be put in 3 ways.

So total number of ways = 3 x 3 x 3 x 3 = 3⁴ = 81

Metric Conversion Charts

Approximate Conversions To Metric Measures

Symbol	When you Know	Multiply by	To Find	Symbol
LENGTH
in ft yd mi	inches feet yards miles	2.5 30.0 0.9 1.6	centimeters centimeters meters kilometers	cm cm m km
AREA
in² ft² yd² mi²	square inches square feet square yards square miles acres	6.5 0.09 0.8 2.6 0.4	square centimeters square centimeters square meters square kilometers hectares	cm² cm² m² km² ha
MASS
oz lb	ounces pounds short tons (2000 lb)	28 0.45 0.9	grams kilograms metric ton	g kg t
VOLUME
tsp Tbsp in³ fl oz c pt qt gal ft³ yd³	teaspoons tablespoons cubic inches fluid ounces cups pints quarts gallons cubic feet cubic yards	5 15 16 30 0.24 0.47 0.95 3.8 0.003 0.76	milliliters milliliters milliliters milliliters liters liters liters liters cubic meters cubic meters	mL mL mL mL L L L L m³ m³
TEMPERATURE (exact)
F	Farenheit Degrees	subtract 32, multiply by 5/9	Celsius Degrees	C

Approximate Conversions from Metric Measures

Symbol	When you Know	Multiply by	To Find	Symbol
LENGTH
mm cm m m km	millimeters centimeters meters meters kilometers	0.004 0.4 3.3 1.1 0.6	inches inches feet yards miles	in in ft yd mi
AREA
cm² m² km² ha	square centimeters square meters square kilometers hectares (10,000 m²)	0.16 1.2 0.4 2.5	square inches square yards square miles acres	in² yd² mi²
MASS
g kg t	grams kilograms metric ton (1,000 kg)	0.035 2.2 1.1	ounces pounds short tons	oz lb
VOLUME
mL mL L L L m³ m³	milliliters milliliters liters liters liters cubic meters cubic meters	0.03 0.06 2.1 1.06 0.26 35.0 1.3	fluid ounces cubic inches pints quarts qallons cubic feet cubic yards	fl oz in³ pt qt gal ft³ yd³
TEMPERATURE (exact)
C	Celsius Degrees	multiply by 5/9, add 32	Farenheit Degrees	F

Approximate Liquid and Dry Measure Equivalencies

Customary	Metric
1/4 teaspoon	1.25 milliliters
1/2 teaspoon	2.5 milliliters
1 teaspoon	5 milliliters
1 tablespoon	15 milliliters
1 fluid ounce	30 milliliters
1/4 cup	60 milliliters
1/3 cup	80 milliliters
1/2 cup	120 milliliters
1 cup	240 milliliters
1 pint (2 cups)	480 milliliters
1 quart (4 cups, 32 ounces)	960 milliliters (0.96 liters)
1 gallon (4 quarts)	3.84 liters
1 ounce (by weight)	28 grams
1/4 pound (4 ounces)	114 grams
1 pound (16 ounces)	454 grams
2.2 pounds	1 Kilogram (1,000 grams)

Oven Temperature Equivalencies

Description	Farenheit	Celsius
Cool	200	90
Very Slow	250	120
Slow	300-325	150-160
Moderately Slow	325-350	160-180
Moderate	350-375	180-190
Moderately Hot	375-400	190-200
Hot	400-450	200-230
Very Hot	450-500	230-260

Cooking Measurement Equivalents

The information below shows measuring equivalents for teaspoons, tablespoons, cups, pints, fluid ounces, and more. This page also includes the conversions for metric and U.S. systems of measurement.

1 tablespoon (tbsp) =	3 teaspoons (tsp)
¹/16 cup =	1 tablespoon
¹/8 cup =	2 tablespoons
¹/6 cup =	2 tablespoons + 2 teaspoons
¹/4 cup =	4 tablespoons
¹/3 cup =	5 tablespoons + 1 teaspoon
³/8 cup =	6 tablespoons
¹/2 cup =	8 tablespoons
²/3 cup =	10 tablespoons + 2 teaspoons
³/4 cup =	12 tablespoons

1 cup =	48 teaspoons
1 cup =	16 tablespoons
8 fluid ounces (fl oz) =	1 cup
1 pint (pt) =	2 cups
1 quart (qt) =	2 pints
4 cups =	1 quart
1 gallon (gal) =	4 quarts
16 ounces (oz) =	1 pound (lb)
1 milliliter (ml) =	1 cubic centimeter (cc)
1 inch (in) =	2.54 centimeters (cm)

Source: United States Dept. of Agriculture (USDA).

U.S.–Metric Cooking Conversions

U.S. to Metric

Capacity		Weight
¹/5 teaspoon	1 milliliter	1 oz	28 grams
1 teaspoon	5 ml	1 pound	454 grams
1 tablespoon	15 ml
1 fluid oz	30 ml
¹/5 cup	47 ml
1 cup	237 ml
2 cups (1 pint)	473 ml
4 cups (1 quart)	.95 liter
4 quarts (1 gal.)	3.8 liters

Metric to U.S.

Capacity		Weight
1 milliliter	¹/5 teaspoon	1 gram	.035 ounce
5 ml	1 teaspoon	100 grams	3.5 ounces
15 ml	1 tablespoon	500 grams	1.10 pounds
100 ml	3.4 fluid oz	1 kilogram	2.205 pounds = 35 ounces
240 ml	1 cup
1 liter	34 fluid oz = 4.2 cups = 2.1 pints = 1.06 quarts = 0.26 gallon

Scale Factors - I

or 'Why There Are No Giant Spiders!

Giant creatures the stuff of many science fiction movies ... from giant ants and spiders to 10-storey tall babies. But is it really possible for creatures to be so large? If so, why aren't there any 6 metre tall spiders?

On this page we'll try to explain why ants and spiders could never be as big as an elephant ... and still look the same.

The reason for this has to do with scale factors ...

Here's a simple box. Beside it is one twice as big.
You can tell it's twice as big because we've put a scale beside it.

Actually, to be more precise, the second box is twice as long as the first one.
The correct way to say this is that it has been scaled up by a factor of two.

By making the box twice as long, we've also doubled the width and height.

Here's the same diagram, only this time we've shaded one face.
The second box is still twice as long as the first one.

Notice what happens to the area of one face of the box when it's twice as long ...

The AREA has been increased by a factor of FOUR.

Here's the same diagram again, only this time we've shaded all the boxes.
The second box is still twice as long as the first one.

Notice what happens to the volume of the box when it's twice as long ...

The VOLUME has been increased by a factor of EIGHT.

Increasing the length of an object by a factor of 2 increases the area by a factor of 4 and the volume by a factor of 8.

Let's start over with another set of boxes.

Here's a simple box. Beside it is one three times as big.
You can tell it's three times as big because we've put a scale beside it.

To be more precise, the second box is three times as long as the first one.
The correct way to say this is that it has been scaled up by a factor of three.

By making the box three times as long, we've also tripled the width and height.

Here's the same diagram, only this time we've shaded one face.
The second box is still three times as long as the first one.

Notice what happens to the area of one face of the box when it's three times as long ...

The AREA has been increased by a factor of NINE.

Here's the same diagram again, only this time we've shaded all the boxes.
The second box is still three times as long as the first one.

Notice what happens to the volume of the box when it's three times as long ...

The VOLUME has been increased by a factor of TWENTY-SEVEN.

Increasing the length of an object by a factor of 3 increases the area by a factor of 9 and the volume by a factor of 27.

Do you see the pattern? Let's summarize what happened (and add a few more examples) in a table:

	LENGTH	AREA	VOLUME
Scale Factor 2	2 times	4 times	8 times
Scale Factor 3	3 times	9 times	27 times
Scale Factor 4	4 times	16 times	64 times
Scale Factor 5	5 times	25 times	125 times
Scale Factor X	X times	X² times	X³ times

Increasing the length of an object by factor X increases the area by factor X² and the volume by factor X³

Scale Factors -II

Page Two
The reason that scale factors are important is that some physical properties of bodies depend on area, and some depend on volume. This has important consequences for the sizes of those bodies, and how they function.

Mass

The mass of an object depends on its volume. For example, if you increase an object's volume by eight times, its mass will also be eight times greater.

Here's a spider that has grown in size by a factor of three. The big one is three times as long as the small one.

From the previous page, you know that this will give the bigger spider a volume that is 27 times the volume of the small spider. This means the bigger spider's mass is 27 times the mass of the smaller one. The big one is 27 times heavier.

This principle holds true whenever you enlarge someting. If a dog were to grow to 5 times its normal size, it would weigh 125 times as much. The weight of an object increases as the cube of the scale factor. Another example? Suppose an ant were to grow 200 times bigger. It would weigh 200³, or 8,000,000 times as much.

Strength

The strength of an organism depends on its cross-sectional area. For example, you might measure your strength by how much weight you could lift with your biceps muscle. If you wanted to get stronger, you would exercise that muscle to make it bigger. But what will increase is the cross-sectional area of the muscle. It will get wider in the middle, but won't get longer.

For example, a muscle that is twice as wide might be four times stronger.
The strength of an organism increases as the square of the scale factor.
Another example? A leg that is twice as big will support four times as much weight. An ant that is 200 times bigger will be 200², or 40,000 times stronger.

Do you see the problem now? As an organism gets bigger, its weight increases as the cube of the scale factor, but its strength increases only as the square. Its strength isn't increasing as fast as its weight.

Let's use an actual example to make it clear what's happening. We'll use the example of an ant that has grown 200 times in length.

Let's make the normal ant's strength 1 unit, and its weight 1 unit. We can then make a ratio of strength to weight by dividing these numbers. We get 1/1 = 1. We'll call this ratio the relative strength ... for the normal ant it's 1.

Now let's work it out for the big ant.
The big ant is 200 times as long, so it will be 200², or 40,000 times stronger.
The big ant's weight is 200³, or 8,000,000 times heavier.
The big ant's strength to weight ratio is 40,000/8,000,000 = 0.005, so it's relative strength is 0.005, or 1/200^th.

Although the ant's size has increased, its strength hasn't kept pace with its weight. It has only 1/200^th of the strength needed to support itself. The result would be an giant ant that couldn't stand up! In fact, since its body shell strength won't have increased enough either, it would end up as a squashed puddle on the floor, totally unable to support itself or its internal organs.

Oxygen Requirements

Cells in living organisms require oxygen to function. If all the cells in an organism were to grow by a factor of X, their surface area would now be X² times as large, but their volume and mass would be X³ times bigger.

The amount of oxygen required depends on the mass of the cells, so the bigger cells would need X³ as much oxygen. However, oxygen gets into the cells through their surface membranes, which have only increased by a factor of X².

The bigger cells' oxygen-aquiring ability won't have increased as fast as their need for oxygen ... so all the cells will die. An organism that increases in size will suffocate!

You may be wondering how any large animals can exist at all, given what we've shown you. For example, when a young animal grows to adulthood, how do its bigger cells breathe?
The answer is simple ... the cells don't get bigger ... they divide. As the organism grows, the number of cells grows, but the cell size remains about the same.

Consider animals that are really large, like the elephant. What do you notice about its body? An elephant isn't shaped like a dog or mouse that's been expanded many times. It's legs are much thicker, in proportion to its body size, than smaller animals.

In order for an organism to be this big, its supporting bones and muscles in the legs must be bigger in proportion to the rest of its body than organisms that are small. Otherwise it wouldn't be strong enough to stand up.

If a spider were as big as an elephant, it would have to look something like this. Notice the very thick legs ... far thicker in proportion to its body than a regular-sized spider.

But of course this spider could never exist. Not only would it suffocate because its cells couldn't get enough oxygen, but it would also be unable to breathe because, although its shell would have to be proportionately thicker to support its inner organs, its area would have only increased as the square of the scale factor, while its mass will have increased as the cube of that value. And a spider breathes through its skin.

So despite what you see in really bad movies, if anyone ever manages to come up with a device to instantly make insects a hundred times bigger, or to enlarge a baby to skyscraper height, none of these enlarged organisms could ever survive. They'd be unable to stand up, and they'd suffocate!

How to find the surface area and volume of prisms.

Theory:

A 'prism' is a solid figure with a uniform cross section. Here are some examples of prisms:

Rectangular based prism:

Circular based prism: (Cylinder)

Triangular based prism:

The surface area of any prism equals the sum of the areas of its faces, which include the floor, roof and walls. Because the floor and the roof of a prism have the same shape, the surface area can always be found as follows:

From now on we will call the 'floor' the 'base'. The height of the prism = 'H'.

The surface area of a prism = 2 × area of base + perimeter of base × H

The actual formula used to find the surface area will depend on the shape of the base of the prism.

For example:

Rectangular based prism Base shape: Rectangle, length 'L' and width 'W' Area of base: L × W Perimeter of base: 2(L+W) Surface area = 2LW + 2(L+W)H

Circular based prism Base shape: Circle, radius 'R' Area of base: pR² Perimeter of base: 2pR Surface area = 2pR² + 2pRH

Triangular based prism Base shape: Triangle: base 'b', height 'h', and sides S₁, S₂ and S₃ Area of base: ½b×h Perimeter of base: S₁+ S₂ + S₃ Surface area = bh + (S₁+ S₂ + S₃)H

Surface Area Formulas

in general, the surface area is the sum of all the areas of all the shapes that cover the surface of the object.

Surface Area of a Cube = 6 a²

(a is the length of the side of each edge of the cube)

In words, the surface area of a cube is the area of the six squares that cover it. The area of one of them is a*a, or a² . Since these are all the same, you can multiply one of them by six, so the surface area of a cube is 6 times one of the sides squared.

Surface Area of a Rectangular Prism = 2ab + 2bc + 2ac

(a, b, and c are the lengths of the 3 sides)

In words, the surface area of a rectangular prism is the are of the six rectangles that cover it. But we don't have to figure out all six because we know that the top and bottom are the same, the front and back are the same, and the left and right sides are the same.

The area of the top and bottom (side lengths a and c) = a*c. Since there are two of them, you get 2ac. The front and back have side lengths of b and c. The area of one of them is b*c, and there are two of them, so the surface area of those two is 2bc. The left and right side have side lengths of a and b, so the surface area of one of them is a*b. Again, there are two of them, so their combined surface area is 2ab.

Surface Area of Any Prism

(b is the shape of the ends)

Surface Area = Lateral area + Area of two ends

(Lateral area) = (perimeter of shape b) * L

Surface Area = (perimeter of shape b) * L+ 2*(Area of shape b)

Surface Area of a Sphere = 4 pi r²

(r is radius of circle)

Surface Area of a Cylinder = 2 pi r² + 2 pi r h

(h is the height of the cylinder, r is the radius of the top)

Surface Area = Areas of top and bottom +Area of the side

Surface Area = 2(Area of top) + (perimeter of top)* height

Surface Area = 2(pi r²) + (2 pi r)* h

In words, the easiest way is to think of a can. The surface area is the areas of all the parts needed to cover the can. That's the top, the bottom, and the paper label that wraps around the middle.

You can find the area of the top (or the bottom). That's the formula for area of a circle (pi r²). Since there is both a top and a bottom, that gets multiplied by two.

The side is like the label of the can. If you peel it off and lay it flat it will be a rectangle. The area of a rectangle is the product of the two sides. One side is the height of the can, the other side is the perimeter of the circle, since the label wraps once around the can. So the area of the rectangle is (2 pi r)* h.

Add those two parts together and you have the formula for the surface area of a cylinder.

Surface Area = 2(pi r²) + (2 pi r)* h

Tip! Don't forget the units.

These equations will give you correct answers if you keep the units straight. For example - to find the surface area of a cube with sides of 5 inches, the equation is:

Surface Area = 6*(5 inches)²

= 6*(25 square inches)

= 150 sq. inches

Special Right Triangles 45º-45º-90º

Certain triangles possess "special" properties that allow us to use "short cut formulas" in arriving at information about their measures. These formulas let us arrive at the answer very quickly.

One such triangle is the 45º-45º-90º triangle.

There are two "special" formulas that apply ONLY to the 45º-45º-90º triangle.

45º-45º-90º (Isosceles Right Triangle)
"Special" Formulas

You must remember that these formulas can be used ONLY in a 45º-45º-90º triangle.

What if I forget the formulas?
What should I do?

The nice thing about mathematics is that there is always another way to do the problem. If you forget these formulas, you could always use the Pythagorean Theorem or a Trigonometry formula.

Let's look at 3 solutions to this problem where you are asked to find x:

Special Formula solution

Pythagorean Theorem solution

Trigonometric solution

We are looking for the hypotenuse so we will use the formula that will give the answer for the hypotenuse:

Substituting the leg = 7, we arrive at the answer:

A nice feature of these special formulas is that the answer is already in reduced form.

Since a 45º-45º-90º, also called an isosceles right triangle, has two legs equal, we know that the other leg also has a length of 7 units.
c²= a²+ b²
x² = 7² +7²x² = 49 + 49
x² = 98