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.
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.
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For example, what order could 16 pool balls be in?
After choosing, say, number "14" we can't choose it again.
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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"
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The factorial function (symbol: !) just means to multiply a series of descending natural numbers. Examples:
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| 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 ...
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The formula is written:
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| 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
3! (10 – 3)!3 × 2 × 1
Binomial
A binomial is a polynomial with two terms
example of a binomial
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| 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.
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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:
an-kbk
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!
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!
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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)
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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:
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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
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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:
Let's see if the formula works:
Yes, it works! Try another value for yourself.
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
