Finally !

Heyy guys, this is Sargam! We all are very close to an end of this course and this feeling is also not great because we all will have to say goodbye to Mr.Piatek !! :(

Well, moving on, today we mainly discussed about the review and some really important questions of Permutation, combination and binomial theorem. Today Mr.Piatek reminded us to do some the entity and some really tricky questions which are very much likely to be on the exam.

Permutations without Repetition

In this case, we have to reduce the number of available choices each time.

For example, what order could 16 pool balls be in? After choosing, say, number "14" we can't choose it again.

So, our first choice would have 16 possibilites, and our next choice would then have 15 possibilities, then 14, 13, etc. And the total permutations would be:

16 × 15 × 14 × 13 × ... = 20,922,789,888,000

But maybe we don't want to choose them all, just 3 of them, so that would be only:

16 × 15 × 14 = 3,360

In other words, there are 3,360 different ways that 3 pool balls could be selected out of 16 balls.

But how do we write that mathematically? Answer: we use the "factorial function"

	The factorial function (symbol: !) just means to multiply a series of descending natural numbers. Examples: 4! = 4 × 3 × 2 × 1 = 24 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5,040 1! = 1
Note: it is generally agreed that 0! = 1. It may seem funny that multiplying no numbers together gets us 1, but it helps simplify a lot of equations.

So, if we wanted to select all of the billiard balls the permutations would be:

16! = 20,922,789,888,000

But if we wanted to select just 3, then we have to stop the multiplying after 14. How do we do that? There is a neat trick ... we divide by 13! ...

16 × 15 × 14 × 13 × 12 ...		= 16 × 15 × 14 = 3,360
13 × 12 ...

The formula is written:

where n is the number of things to choose from, and we choose r of them (No repetition, order matters)

Examples:

Our "order of 3 out of 16 pool balls example" would be:

16!	=	16!	=	20,922,789,888,000	= 3,360
(16-3)!		13!		6,227,020,800

(which is just the same as: 16 × 15 × 14 = 3,360)

How many ways can first and second place be awarded to 10 people?

10!	=	10!	=	3,628,800	= 90
(10-2)!		8!		40,320

(which is just the same as: 10 × 9 = 90)

Combinations

The number of ways of selecting r objects from n unlike objects is:

Example

There are 10 balls in a bag numbered from 1 to 10. Three balls are selected at random. How many different ways are there of selecting the three balls?

¹⁰C₃ =10!=10 × 9 × 8= 120
             3! (10 – 3)!3 × 2 × 1

Binomial

A binomial is a polynomial with two terms

example of a binomial

Multiplying

The Binomial Theorem shows what happens when you multiply a binomial by itself (as many times as you want).

It works because there is a pattern ... let us see if we can discover it.

Exponents

But first you need to know what an Exponent is. Here is a quick summary: An exponent says how many times to use something in a multiplication.

Example: 8² = 8 × 8 = 64

An exponent of 1 means just to have it appear once, so you get the original value:

Example: 8¹ = 8

An exponent of 0 means not to use it at all, and we have only 1:

Example: 8⁰ = 1

Exponents of (a+b)

Now on to the binomial.

We will use the simple binomial a+b, but it could be any binomial.

Let us start with an exponent of 0 and build upwards.

Exponent of 0

When an exponent is 0, you get 1:

(a+b)⁰ = 1

Exponent of 1

When the exponent is 1, you get the original value, unchanged:

(a+b)¹ = a+b

Exponent of 2

An exponent of 2 means to multiply by itself (see how to multiply polynomials):

(a+b)² = (a+b)(a+b) = a² + 2ab + b²

Exponent of 3

For an exponent of 3 just multiply again:

(a+b)³ = (a+b)(a² + 2ab + b²) = a³ + 3a²b + 3ab² + b³

We have enough now to start talking about the pattern.

The Pattern

In the last result we got:

a³ + 3a²b + 3ab² + b³

Now, notice the exponents of a. They start at 3 and go down: 3, 2, 1, 0:

Likewise the exponents of b go upwards: 0, 1, 2, 3:

If we number the terms 0 to n, we get this:

k=0	k=1	k=2	k=3
a3	a2	a	1
1	b	b2	b3

Which can be brought together into this:

a^n-kb^k

How about an example to see how it works:

Example: When the exponent, n, is 3.

The terms are:

k=0:	k=1:	k=2:	k=3:
an-kbk = a3-0b0 = a3	an-kbk = a3-1b1 = a2b	an-kbk = a3-2b2 = ab2	an-kbk = a3-3b3 = b3

It works like magic!

Coefficients

So far we have:	a3 + a2b + ab2 + b3
But we really need:	a3 + 3a2b + 3ab2 + b3

We are missing the numbers (which are called coefficients).

Let's look at all the results we got before, from (a+b)⁰ up to (a+b)³:

And now look at just the coefficients (with a "1" where a coefficient wasn't shown):

They actually make Pascal's Triangle!

Each number is just the two numbers above it added together (except for the edges, which are all "1")

(Here I have highlighted that 1+3 = 4)

Armed with this information let us try something new ... an exponent of 4:

a exponents go 4,3,2,1,0:		a4	+	a3	+	a2	+	a	+	1
b exponents go 0,1,2,3,4:		a4	+	a3b	+	a2b2	+	ab3	+	b4
coefficients go 1,4,6,4,1:		a4	+	4a3b	+	6a2b2	+	4ab3	+	b4

And that is the correct answer. 

As a Formula

Our last step is to write it all as a formula.

But hang on, how do we write a formula for "find the coefficient from Pascal's Triangle" ... ?

Well, there is such a formula:

It is commonly called "n choose k" because it is how many ways to choose k elements from a set of n. You can read more at Combinations and Permutations The "!" means "factorial", for example 4! = 1×2×3×4 = 24

And it matches to Pascal's Triangle like this: (Note how the top row is row zero and also the leftmost column is zero!)

Example: Row 4, term 2 in Pascal's Triangle is "6".

Let's see if the formula works:

Yes, it works! Try another value for yourself.

Notes: 

1. A formula can be used only when there is no restriction. 

2. If finding the number of terms in an expansion , the number is terms are always one number bigger than the exponent. For example: (a+b)^20 =  the number of terms would be 21 

3. When finding the entity or the number of ways A and B must sit together, first find the total permutations, then calculate when a and b are together and subtract the result from the total permutation calculation result. 

That would be all I guess, All the very best to everybody for their exams and to those to who are currently having their English exams....! :) 

“If you want to achieve excellence, you can get there today. As of this second, quit doing less-than-excellent work.”- Thomas J Watson

RADICAL AND RATIONAL FUNCTIONS UNIT 7

Hi Mr.Piateks pre-cal 40S class! My name is simran and I'm going to be giving an explanation on the radical and rational functions unit. Sorry i couldn't post any pictures of graphs due to problems with my computer, however, i hope the explanation is beneficial to you all! Good luck for the exam!

Example 1: Graphing a Radical Function

1.             Fill in the table of values for the function f(x) = root(x).

2.             When constructing a table of values for a square root function, it's a good

idea to select x-values that are perfect squares to avoid decimals.

3.             The square root of -1 is undefined.

4.             The square root of zero is 0.

5.             The square root of 1 is 1.

6.             The square root of 4 is 2.

7.             And the square root of 9 is 3.

8.             Draw the graph of the function f(x) = root(x) and state the domain and range.

9.             The domain is {x|x ≥ 0,xER}

10.    The range is {y|y ≥ 0,yER}

11.    State that the Radicand is greater than or equal to 0

12.    The radicand is x greater than or equal to 0

13.    State the endpoint and the x and y intercepts

14.    The endpoint is @ (0,0)

15.    The x-intercept @ x=0 and the y-intercept @ y=0

 NOTE: when graphing radical functions using multiple transformations graph in steps, starting with the basic function y= root(x) then horizontal stretch, horizontal reflection in the y-axis, vertical stretch, vertical reflection in the x-axis, and lastly translations.

Example 2: Reflections of Radical Functions

1.             Graph the function of y = root(x) as a reference. In Part A, draw the graph of f(x) = -root(x).

2.             A negative coefficient tells you to reflect the graph about the x-axis.

3.             Reflect the original graph about the x-axis to get the graph of f(x) = -root(x).

4.             In Part B, draw the graph of f(x) = root(-x).

5.             A negative coefficient within the radical tells you to reflect the graph about the y-axis.

6.             Reflect the original graph about the y-axis to get the graph of f(x) = root(-x).

Example 3: Stretches of Radical Functions

1.             Graph the function of y = root(x) as a reference. In Part A, draw the graph of f(x) = 2root(x).

2.             The coefficient of the radical is the vertical stretch of the graph.

3.             Double the height of the graph to draw f(x) = 2root(x).

4.             In Part B, draw the graph of f(x) = 1/2root(x).

5.             The coefficient of the radical is the vertical stretch of the graph.

6.             Halve the height of the graph to draw f(x) = 1/2root(x).

7.             In Part C, draw the graph of f(x) = root(2x).

8.             The reciprocal of the internal coefficient is the horizontal stretch of the graph.

9.             The reciprocal of 2 is 1/2, so the width of the graph is halved.

10.          In Part D, draw the graph of f(x) = root(1/2x).

11.          The reciprocal of the internal coefficient is the horizontal stretch of the graph.

12.          The reciprocal of 1/2 is 2. The width of the graph is doubled.

Example 4: Translations of Radical Functions

1.             Graph the graph of y = root(x) as a reference. In Part A, draw the graph of f(x) = root(x) - 5.

2.             The number added or subtracted from the radical is the vertical translation of the graph.

3.             Move the original graph down 5 units to get the graph of f(x) = root(x) – 5.

4.             In Part B, draw the graph of f(x) = root(x) + 2.

5.             The number added or subtracted from the radical is the vertical translation of the graph.

6.             Move the original graph up 2 units to get the graph of f(x) = root(x) + 2.

7.             In Part C, draw the graph of f(x) = root(x – 1).

8.             The OPPOSITE of this number is the horizontal translation of the graph.

9.             The opposite of -1 is +1. Move the original graph 1 unit right.

10.          In Part D, draw the graph of f(x) = root(x + 7).

11.          The OPPOSITE of this number is the horizontal translation of the graph.

12.          The opposite of +7 is -7. Move the original graph 7 units left.

Example 5: Multiple Transformations of Radical Functions

1.             Graph the function of y = root(x) as a reference. In Part A, draw the graph of f(x) = root(x - 3) + 2.

2.             There is a horizontal translation of three units right.

3.             There is also a vertical translation of two units up.

4.             Move the graph 3 units right and 2 units up to get the graph of f(x) = root(x - 3) + 2.

5.             In Part B, draw the graph of f(x) = 2root(x + 4).

6.             There is a vertical stretch by a scale factor of 2.

7.             There is also a horizontal translation of 4 units left.

8.             Double the height of the graph and move it 4 units left. Note that stretches and reflections must be performed before translations.

9.             In Part C, draw the graph of f(x) = -root(x) - 3.

10.          There is a reflection about the x-axis.

11.          There is a vertical translation 3 units down.

12.          Reflect the graph across the x-axis and move it 3 units down. Note that stretches and reflections must be performed before translations.

13.          In Part D, draw the graph of f(x) = root(-2x - 4).

14.          Before you graph this radical function, you need to make sure the binomial within the radical sign is fully factored.

15.          Factor out -2 to get f(x) = root(-2(x + 2)).

16.          The graph is horizontally stretched by a factor of 1/2, reflected about the y-axis, and horizontally translated 2 units left.

NOTE: to find the x-intercept plug in 0 for y into the original equation and solve for x.

            to find the y-intercept plug in 0 for x into the original equation and solve for y.

DETERMINING RADICAL FUNCTIONS

When determining a radical function from a graph the very first step is to draw the graph of y=root(x). Then you can compare the basic graph with the transformed graph.

Determining the a value: examine the graph and detect two x values that are the same for both of the functions (basic function and transformed function). Determine the y values with the corresponding x values. Making a table of values with the basic function on one side and transformed function on the other side in order to compare your values is recommended. Determine the difference between the two y-values from the two different functions. The difference is the a value

Determining the b value: examine the graph and detect two y values that are the same for both of the functions (basic function and transformed function). Determine the x values with the corresponding y values. Determine the difference between the two x-values from the two different functions. The difference is the b value.