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=**Welcome to Your Class Wiki! **= PLEASE CITE YOUR MATERIAL GRADES:

1. ** __Definition:__ If One Event Occurs In (M) Ways And Another Occur In (N) Ways, Then The Number Of Ways Both Can Occur Is (MxN) __Ex.1__ Your Buying A Pizza. You Have A Choice Of 3 Crusts, 4 Cheeses, 5 Vegtables, And 8 Meat Toppings. How Many Different Pizzas With 1 Crust, 1 Cheese, 1 Meat, And 1 Vegtable Can You Choose? __How To Solve:__ 3x4x8x5= 480**
 * I. __Fundamental Counting Principles:__ **
 * [] << Resource**


 * __Ex.2__** Arranging the letters in the name "FRED"
 * __How To Solve:__** There are four ways to pick the first letter, then there are three ways to choose the second letter, and then two ways to choose the third letter. 4x3x2= 24 arrangements.


 * __F__ ||  || __FR__ ||   || __FRE__ || __FRED__ ||
 * __FRD__ || __FRDE__ ||  ||
 * __FE__ ||  || __FER__ || __FERD__ ||
 * __FED__ || __FEDR__ ||  ||
 * __FD__ ||  || __FDE__ || __FDER__ ||
 * __FDR__ || __FDRE__ ||  ||   ||
 * __R__ ||  || __RF__ ||   || __RFE__ || __RFED__ ||
 * __RFD__ || __RFDE__ ||  ||
 * __RE__ ||  || __REF__ || __REFD__ ||
 * __RED__ || __REDF__ ||  ||
 * __RD__ ||  || __RDF__ || __RDFE__ ||
 * __RDE__ || __RDEF__ ||  ||   ||
 * __E__ ||  || __EF__ ||   || __EFR__ || __EFRD__ ||
 * __EFD__ || __EFDR__ ||  ||
 * __ER__ ||  || __ERF__ || __ERFD__ ||
 * __ERD__ || __ERDF__ ||  ||
 * __ED__ ||  || __EDF__ || __EDFR__ ||
 * __EDR__ || __EDRF__ ||  ||   ||
 * __D__ ||  || __DF__ ||   || __DFR__ || __DFRE__ ||
 * __DFE__ || __DFER__ ||  ||
 * __DR__ ||  || __DRF__ || __DRFE__ ||
 * __DRE__ || __DREF__ ||  ||
 * __DE__ ||  || __DEF__ || __DEFR__ ||
 * __DER__ || __DERF__ ||  ||   ||

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**2**. __**Different Methods**__ 1. __**Factorials**__: The number of ways to arrange ** all ** of a distinct number of items, symbolized by ** n! **, multiplying down beginning from the cardinal number n, all the way to 1. [] [] << resource 2.**__Tree Diagrams__**: A visual describing all the possible outcomes of a multi-part task. **__Steps__:** • Use the Fundamental Counting Principle to figure out the number of outcomes and then set up the tree diagram using each part of the task as headings. list the possible outcomes under each heading. Each possible outcome from the first heading will create a branch to each possible outcome from the next heading. List all the possible outcomes (sample points) under the heading, // Sample Space. // __**Ex.1**__ Ruth is playing a game where she flips a coin and then rolls a die. []
 * __Ex.1__** Six students are to occupy 6 desks in a row. How many ways are there to arrange them?
 * __Problem__**: 6!=6x5x4x3x2x1
 * __Answer__**: 720
 * __Ex.2__** [[image:http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/series/gifs/convergence_28.gif]]
 * __Problem__**: 6x2
 * __Answer__**: 12 coin die []


 * 3.** **__Permutations__**: An **ordered** arrangement of a given set of items determined by choosing a distinct qauntity of item
 * (r) ** from a certain total of items ( ** n ** ). ** Formula: nPr = n! **(n-r)! //Use this method when: •Repetition is// ** //not// **// permitted • Order ** is **// //important//

and Vice-President (2 officers)? 4. **__Combinations__**: given set of items ** without ** regard to their arrangement determined by selecting a distinct quantity of items (** r ** ) from a certain total of items ( ** n ** ). Formula: nCr = n! (n-r)!r!
 * __Ex.1__**//:// If a club has 7 members, how many different ways are there to choose a President
 * __Ex.1__**//:// If a club has 7 members, how many different ways are there to choose a President
 * __Problem__**: 7! = 7! = 7•6•5•4•3•2•1 = 42
 * __Answer__**: (7-2)! 5x4x3x2x1

media type="youtube" key="eEC1f97XYGY" height="385" width="480"

you use Use this method when: • Repetition ** is not ** permitted • Order ** is not ** important (i.e. subsets, committees, random selections, card hands)(When the set of {a,b,c} is the same as a, b, and c. __**Ex.1:**__ If a club has 7 members, how many different ways are there to choose a 2 member sub-committee? 7 C 2 = 7! 7•6•5•4•3•2•1 42
 * __Problem__**: (7-2)! 2! = 5•4•3•2•1•2•1 = 2
 * __Answer__**: 21 []

If something can be chosen, or can happen, in //m// different ways, and, //after that has happened//, something else can be chosen in //n// different ways, then the number of ways of choosing both of them is //m// **·** //n//. For example, imagine putting the letters //a, b, c, d// into a hat, and then drawing two of them one after another. We can draw the first in 4 different ways: either //a// or //b// or //c// or //d//. After that has happened, there are 3 ways to choose the second. That is, after you have chosen one of the 4 ways, there are only 3 other ways to choose another letter. Therefore, there are 4**·** 3 or 12 possible ways to choose two letters from four.
 * //ab// ||  ||   ||   || //ba// ||   ||   ||   || //ca// ||   ||   ||   || //da// ||




 * //ac// ||  ||   ||   || //bc// ||   ||   ||   || //cb// ||   ||   ||   || //db// ||




 * //ad// ||  ||   ||   || //bd// ||   ||   ||   || //cd// ||   ||   ||   || //dc// ||

//ab// means that //a// was chosen first and //b// second; //ba// means that //b// was chosen first and //a// second; and so on. Let us now consider the total number of permutations of all four letters. There are 4 ways to choose the first. 3 ways remain to choose the second, 2 ways to choose the third, and 1 way to choose the last. Therefore the number of permutations of 4 different things is 4**·** 3**·** 2**·** 1 = 24 The number of permutations of 4 different things taken 4 at a time is 4!. ("4! is the number of permutations of 4 different things from a total of 4 different things.") In general, The number of permutations of n different things taken n at a time is n!. []

Circular Permutation - When things are arranged in places along a closed curve or a circle, in which any place may be regarded as the first or last place, they form a **circular permutation**

The following arrangements of 4 objects O1, O2, O3, O4 in a circle will be considered as one or same arrangement.



Observe carefully that when arranged in a row, O1 O2 O3 O4, O2O3O4 O1, O3O4O1O2, O4O1O2O3 are different permutations. When arranged in a circle, these 4 permutations are considered as one permutation.The number of circular permutations of n different objects is (n-1)!.Each circular permutation corresponds to n linear permutations depending on where we start. Since there are exactly n! linear permutations, there are exactly **n!/n** permutations. Hence, the number of circular permutations is the same as (n-1)!.

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Two types of permutation__: 1. Repetition is Allowed: such as a pad lock. It could be "333". 2. No Repetition: an example of it would be if the first three people in a running race. You can't be first and second.** []
 * II.** **__Permutations:

1. Permutations With Repetion - When you have **//n//** things to choose from, you have **//n//** choices everytime. So when choosing **//r//** of them, the permutations are, n x n x ... (r times) = n^r So, the formula is simply:


 * **n^r** ||

(Repetition allowed, order matters) ||
 * where //**n**// is the number of things to choose from, and you choose //**r**// of them

Find the number of distinguishable permutations of the letters in each word. //n=// the number of objects //r=// the number of repeats
 * CLIMB**

=5! / 0!= 5! = 120 -Since climb has 5 letters in it, thats what //n// is and because it has no repetition of letters 0 is //r.// =10! / (2!x3!x3!)= 50,400 -Cincinatti has 10 letters and 3 different letters have repeats. C is repeated twice, N is repeated three times, as well as I. So you multiply all the repeats together. Which is why you get (2!x3!x3!). []
 * CINCINATI**

2. Permutation Without repetition The formula is written:


 * n! / (n - r)! ||

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

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** 6. Practice Problems.. **
 * 1.** License plates are being made with three letters and three digits. How many different license plates can be made:

A.) If Letters and digits can be repeated
 * __Answer:__** **10x10x10x26x26x26 = 17,576,000**

B.) If letters and digits cannot be repeated
 * __Answer:__** **10x9x8x26x25x24 = 11,232,000**

**3.** You have 6 homework assignments to complete over the weekend. You only have time to complete 4 of them on saturday. In how many ways can you comlete the 4 assignments?
 * 2. ** Find the distinguishable permutations of the letters in the word ADRIANNA.
 * __Answer:__** **8! / (3!x2!) = 3360**
 * __Answer:__** **6C4 = 6! / (6-4)! x 4! = 8640**

Resourses --> []

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For example, imagine putting the letters //a, b, c, d// into a hat, and then drawing two of them in succession. We can draw the first in 4 different ways: either //a// or //b// or //c// or //d//. After that has happened, there are 3 ways to choose the second. That is, to //each// of those 4 ways there correspond 3. Therefore, there are 4**·** 3 or 12 possible ways to choose two letters from four. i found this from, [Page(n.d.). Retrieved from http://www.themathpage.com/aprecalc/permutations-combinations.htm|Resource Page(n.d.). Retrieved from http://www.themathpage.com/aprecalc/permutations-combinations.htm] ==C!/(n-r)!r! **-A __combination__ is one of the many different arrangements of a group of items where __order does not matter.__ **check out []==
 * 1) **//n= Total # of objects r= Chosen objects []//**
 * 2) **//The number of combinations of r objects taken from the group of n distinct objects is denoted by//**

__n!__
(n-r)! **. r!** Instead of writing the whole formula, people use different notations such as these: is the equation I found at [] ==Proof: Example: **//P//(10,2) = 10! / (10 - 2)! = 90 []** == == ==

**3.** Combinations can be used to determine different ways to arrange items in variety of combinations. __For example:__ Five people are in a club and three are going to be in the 'planning committee,' to determine how many different ways this committee can be created we can use a combination formula as follows: **n C r= n!**  To find all of the differennt ways to arrange r items out of n items. Use the combination formula below. (n stands for the total number of items; r stands for how many things you are choosing.)
 * r!(n-r)!**
 * We can use combination to see how many ways can things arranged.** **For example if a four person group is chosen randomly from a group of 18 people.**



Go to [] for more details. Would I **use combinations** or permutations? If 6 people are running in a race, how many possible ways can they come in if there are no ties **...** [- Cached |http://www.enotes.com/math/q-and-a/would-use-combinations-permutations-50219 - Cached]

**//Example 1//**: To find the number of ways to take 20 and arrange them in 5 groups when order doesn't matter

(which is just the same as: **16 × 15 × 14 = 3360)** []
 * //Example 2 ://** The order of 3 out of 16 pool balls example would be:
 * **16!** || = = || **16!** || = = || **20,922,789,888,000** || **= 3360** ||
 * **(16-3)!** ||^  || **13!** ||^   || **6,227,020,800** ||^   ||
 * **(16-3)!** ||^  || **13!** ||^   || **6,227,020,800** ||^   ||

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4. Mathematical combination can be use to help provide a method to determine answer based on a variety of variation. For example, Fast Food Combinations: How many possible combinations can be made from a special menu of eight items? Several contributors noticed when McDonald's introduced a special menu with 8 different items. An ad claimed that this creates 40312 possible combinations. Is this true? See the answer at [] ======

**__permutation of a number of objects is the number of different ways they can be orderer. With combinations__** __**one does not consider the order in which objects were chosen or placed, just which objects were chosen .**__ I found this at []

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5**. __Pascals Triangle__, is a geometric arrangement of the binomial coefficients in a triangle. It is named after mathematician __Blaise Pascal.__ ​**[]===== A) The triangle was actually discovered in China, the first "pascals" triangle was developed by a mathematician named Hue Yang. []

Pascals triangle was published three years after his death in 1665. The name came from the first westerner to study the triangle. [] B) The triangle is very useful to gamblers, as horse racing, lottery wheels, or British football pools. [] **

Pascals triangle has an important role on combination, raise of powers, and the bionomial theorem. Pascals triangle can also be used to compute a premutation when a mulitplying factor is introduced. Pascals triangle can determine the answer directly of combinations as compared to other branches of mathematics.  [] ​

The answer to both problems is contained in Pascal's Triangle. Modern calculators and spreadsheets embed a simpler way of working out these calculations. In Excel use the combin(n,r) function, where n is the number of articles and r is the repetitions (combinations). you can find this information on []  The Difference between combinations and permutations... ​ the permutation of a number of objects is the number of different ways they can be ordered. With combinations though, it does not matter the order which the objects were chosen, just which objects were chosen.

[] ===To build the triangle, start with "1" at the top, then continue placing numbers below it in a triangular pattern. 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)** [|http://www.mathsisfun.com/pascals-triangle.html] A. The so called 'Pascal' triangle was known in China as early as 1261. Yang Hui applied the triangle to Jia Xian, who lived in the eleventh century'. It was used in the eleventh century for solving quadratic and cubic equations, even, though it seems to have been known since the first millennium. The Pascal triangle known today was corrected and extended it a useable method. There are some proofs that this number triangle was familiar to the Arab astronomer, poet and mathematician Omar Khayyam as early as the 11th century. Most think the triangle came to Europe from China through Arabia. Chinese often equally Pascal`s Triangle as being found in his work that published it for the first time after his death. Both Blaise Pascal and his father Etienne had been in correspondence with Father Marin Mersenne, who published a book with a table of binomial coefficients in 1636. Other authors discussed the idea with respect. Like, expansions of binomials, answers to combinatorial problems and figurate numbers, numbers relating to figures such as triangles, squares, tetrahedra and pyramids. []
 * One of the most interesting Number Patterns is Pascal's Triangle (named after //Blaise Pascal//, a famous French Mathematician and Philosopher).

B. As stated previously, the coefficients of (//x// + 1)//n// are the nth row of the triangle. Now the coefficients of (//x// − 1)//n// are the same, except that the sign alternates from +1 to −1 and back again. After suitable normalization, the same pattern of numbers occurs in the Fourier transform of sin(//x//)//n//+1///x//. More precisely: if //n// is even, take the real part of the transform, and if //n// is odd, take imaginary part. Then the result is a step function, whose values (suitably normalized) are given by the //n//th row of the triangle with alternating signs.[//citation needed//] For example, the values of the step function that results from [|http://en.wikipedia.org/wiki/Pascal's_triangle] ||  ||

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media type="youtube" key="Zo2JrPjijHc" height="385" width="640" Definition : Permutation means //arrangement// of things. The word //arrangement// is used, if the order of things //is considered//. go to [] for more details

Permutation: Suppose we want to find the number of ways to arrange the three letters in the word CAT in different two-letter groups where CA is different from AC and there are no repeated letters.

Because order matters, we're finding the number of permutations of size 2 that can be taken from a set of size 3. This is often written 3_P_2. We can list them as:

The formula is written:  (No repetition, order matters) || go to [] for more information
 * ** ! ** || 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 = 5040
 * 1! = 1 ||
 * Note: it is generally agreed that **0! = 1**. It may seem funny that multiplying no numbers together gets you 1, but it helps simplify a lot of equations. ||
 * where //**n**// is the number of things to choose from, and you choose **//k//** of them

Permutation Examples:
ex. 1 : Our "order of 3 out of 16 pool balls example" would be: (which is just the same as: **16 × 15 × 14 = 3360**) ex 2: How many ways can first and second place be awarded to 10 people? (which is just the same as: **10 × 9 = 90**) [] []
 * **16!** || = = || **16!** || = = || **20,922,789,888,000** || **= 3360** ||
 * **(16-3)!** ||^  || **13!** ||^   || **6,227,020,800** ||^   ||
 * **(16-3)!** ||^  || **13!** ||^   || **6,227,020,800** ||^   ||
 * **10!** || = = || **10!** || = = || **3,628,800** || **= 90** ||
 * **(10-2)!** ||^  || **8!** ||^   || **40,320** ||^   ||
 * **(10-2)!** ||^  || **8!** ||^   || **40,320** ||^   ||

CA CT AC AT TC TA Combination: When we want to find the number of combinations of size 2 without repeated letters that can be made from the three letters in the word CAT, order doesn't matter; AT is the same as TA. We can write out the three combinations of size two that can be taken from this set of size three:

CA CT AT

In combination order does not matter.

Example: "My fruit salad is a combination of apples, grapes and bananas" We don't care what order the fruits are in, they could also be "bananas, grapes and apples" or "grapes, apples and bananas" its the same fruit salad.

6. [] 7.**1. Problem:** **//4 Σ (2k + 1) k = 1// 1. Problem: //5!//

2. Problem: If a school has lockers with 50 numbers on each combination lock, how many possible combinations using three numbers are there.** [] IV. = =

PERMUTATIONS AND COMBINATIONS
In __permutations__, the order is all important (you would //abc// as different from //bca)//. But in __combinations__ we are concerned only that //a, b,// and //c// have been selected .( //abc// and //bca// are the same combination ) choose only three from ABCD,and you can get one of these 4 combinations  // abc abd acd bcd .// The number of combinations of 4 things taken 3 at a time. We will denote this number as 4 //C// 3. In general, (four choose three) //n// //C// //k// = // The number of combinations of n things taken k at a time. // Now, how are the number of combinations //n// //C// //k// related to the number of [|permutations], //n// //P// //k// ? To be specific, how are the combinations 4 //C// 3 related to the permutations 4 //P// 3 ? Since the order does not matter in combinations, there are clearly fewer combinations than permutations. The combinations are contained among the permutations -- they are a "subset" of the permutations. Each of those four [|combinations], in fact, will give rise to 3! permutations: Each column is the 3! permutations of that combination. But they are all //one// combination -- because the order does not matter. Hence there are 3! times as many permutations as combinations. 4 //C// 3, therefore, will be 4 //P// 3 divided by 3! -- the number of //permutations// that each combination generates. 3! || = || __4**·** 3**·** 2__ 1**·** 2**·** 3 || //k//! || . || k//!// || Example 1. How many combinations are there of 5 things taken 4 at a time? 1**·** 2**·** 3**·** 4 || = 5 || Again, both the numerator and denominator have the //number of factors// indicated by the lower index, which in this case is 4. The numerator has four factors beginning with the upper index 5 and going backwards. The denominator is 4!.
 * //abc// ||  || //abd// ||   || //acd// ||   || //bcd// ||
 * //acb// ||  || //adb// ||   || //adc// ||   || //bdc// ||
 * //bac// ||  || //bad// ||   || //cad// ||   || //cbd// ||
 * //bca// ||  || //bda// ||   || //cda// ||   || //cdb// ||
 * //cab// ||  || //dab// ||   || //dac// ||   || //dbc// ||
 * //cba// ||  || //dba// ||   || //dca// ||   || //dcb// ||
 * 4 //C// 3 || = || [[image:http://www.themathpage.com/aPreCalc/Pre_IMG/118.gif width="20" height="22"]]
 * Notice**: The numerator and denominator have the same number of factors, 3, which is indicated by the lower index. The numerator has 3 factors starting with the //upper// index and going down, while the denominator is 3!.
 * In general, //n// //C// //k// = || [[image:http://www.themathpage.com/aPreCalc/Pre_IMG/119.gif width="20" height="22"]]
 * //n// //C// //k// || = || __//n//(//n// − 1)(//n// − 2)**·** **·** **·**__ //__to__// k //factors//
 * **Solution**. 5 //C// 4 = || __5**·** 4**·** 3**·** 2 __

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​​ <span style="font-family: Tahoma,Geneva,sans-serif;">**Example of a Combination:** 1) <span style="color: #800080; font-family: Tahoma,Geneva,sans-serif;">You have a deck of 52 cards. You take out the 4 aces and hold them, but its doesn't matter which order you hold them in. It could be the ace of hearts first, spades, clubs, and diamonds. Or it could be the ace of diamonds first, hearts, clubs, and spades. It doesnt matter which order its in, therefore its a combination. 2) If three letters are chosen from A, B, C, D, E and F then the number of combinations would be: If three letters are chosen from A, B, C, D, E and F then the number of combinations would be: 3!(6-3)! || = || 6! 3!3! || = 20 ||
 * || 6!

<span style="color: #000000; font-family: Tahoma,Geneva,sans-serif;">Example of a Premutation: 1) <span style="color: #800080; font-family: Tahoma,Geneva,sans-serif;">Your locker combination is 0-1-2. In order to open your locker, you have to put in the 0 first, then the 1, then the 2. You cant put in the 2 first then 0 then 1. It has to be in order, therefore its a premutation. 2) given a set of five different objects, when two objects are chosen out of the five. (5 - 2)! || = || 20 Permutations ||
 * || 5!

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V. Creativity figure out the likelihood of something happening.
 * VI.Probability**

Experiment 2: A single 6-sided die is rolled. What is the probability of each outcome? What is the probability of rolling an even number? of rolling an odd number?

The possible outcomes are 1,2,3,4,5,6,7

Experiment 2 illustrates the difference between an outcome and an event. A single outcome of this experiment is rolling a 1, or rolling a 2, or rolling a 3, etc. Rolling an even number (2, 4 or 6) is an event, and rolling an odd number (1, 3 or 5) is also an event.
 * Probabilities: || P(1) || = = ||  || = = ||   ||   ||   ||   ||
 * ||  ||   || total # of sides ||   || 6 ||   ||   ||   ||
 * || P(2) || = = ||  || = = ||   ||   ||   ||   ||
 * ||  ||   || total # of sides ||   || 6 ||   ||   ||   ||
 * || P(3) || = = ||  || = = ||   ||   ||   ||   ||
 * ||  ||   || total # of sides ||   || 6 ||   ||   ||   ||
 * || P(4) || = = ||  || = = ||   ||   ||   ||   ||
 * ||  ||   || total # of sides ||   || 6 ||   ||   ||   ||
 * || P(5) || = = ||  || = = ||   ||   ||   ||   ||
 * ||  ||   || total # of sides ||   || 6 ||   ||   ||   ||
 * || P(6) || = = ||  || = = ||   ||   ||   ||   ||
 * ||  ||   || total # of sides ||   || 6 ||   ||   ||   ||
 * || P(even) || = = ||  || = = ||   || = ||   ||   ||
 * ||  ||   || total # of sides ||   || 6 ||   || 2 ||   ||
 * || P(odd) || = = ||  || = = ||   || = ||   ||   ||
 * ||  ||   || total # of sides ||   || 6 ||   || 2 ||   ||
 * || P(6) || = = ||  || = = ||   ||   ||   ||   ||
 * ||  ||   || total # of sides ||   || 6 ||   ||   ||   ||
 * || P(even) || = = ||  || = = ||   || = ||   ||   ||
 * ||  ||   || total # of sides ||   || 6 ||   || 2 ||   ||
 * || P(odd) || = = ||  || = = ||   || = ||   ||   ||
 * ||  ||   || total # of sides ||   || 6 ||   || 2 ||   ||
 * || P(odd) || = = ||  || = = ||   || = ||   ||   ||
 * ||  ||   || total # of sides ||   || 6 ||   || 2 ||   ||

There are many forms of probability problems. The one i will be discussing is with dies. There are 36 out comes no matter what. Say that you roll a two dice. Your outcome adds up to six. What is the possible numbers on the dice? there are 1 and 5, 2 and 4, 3 and 3. There are three possible outcomes. You need to find the percentage of the possibility of getting the added amount of six. www.mathgoodies.com/lessons/vol6/intro_**probability**.html

For Helpful Videos on Both Permutaions and Combinations Refer to the Links Below:

[] [] [] [] You Can also Visit the Following Website for More Help: []

__//Find the answers to the following://__
 * More Permutation Problems:**



4. In how many ways can 6 people be seated in a row? 5. Seven boys and nine girls are in a club. In how many ways can they elect four different officers designated by A, B, C, and D if: A) A and B must be boys and C and D must be girls? A B C D 7 6 9 8 B) two of the officers must be boys and two of the officers must be girls? There are 30 different ways for this combination
 * __Answer:__ 6! = 6x5x4x3x2x1 = 720 ways**


 * __Resource:__** []

= II. Permutations = 1. The definition of n! is factorial (n)(n-1)(n-2) and so on [] 2. For the notations always put the "f" before the number in parenthesis ex. f (5) [] 3. You can use permutations in any word like DOG you can put them into different sets and list them like this DO DG OD OG GD GO [] 5. Ex.#1 making pizzas- you have three types of crust, thick, thin and regular 4 types of toppings pepperoni, mushrooms, pineapple, and ham and 3 types of cheese American, mozzarella, and Swiss

__**How many one topping cheese pizzas be made?**__ Ex. # 2 -Choosing a new car- you have 4 types of new ford cars Focus, Fusion, Five-Hundred, and Taurus. You have 2 choices of interior for your car Leather and Corduroy. And you have 6 different colors to choose from, Blue, Green, Red, Silver, White, and Gray.
 * __Answer:__** **4P3 = 4! / (4 - 3)! = 24**

Answer:__**
 * __How many Fusions could there be with Leather seats?

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**problems for the students to slove**
1) 7 people take part in a panel discussion. Each person is to shake hands with all of the other participants at the beginning of the discussion. How many handshakes take place? List them all.

2) 8 points are arranged in a circle and each point is joined to each other point by a line. How many lines are needed?

3) A basketball team has 11 players on its roster. Only 5 players can be on the court at one time. How many different groups of 5 players can the team put on the floor?
 * __Answer:__** **11C5 = 11! / (11 - 5)! x 5! = 462**

4) Over the weekend, your family is going on vacation, and your mom is letting you bring your favorite video game console as well as five of your games. How many ways can you choose the five games if you have 12 games in all?

5) Eleven students put their names on slips of paper inside a box. Three names are going to be taken out. How many different ways can the three names be chosen?

[] 1. The idea that if we have //a// ways of doing something and //b// ways of doing something else, then there are //a// x //b// ways of doing both things. Also known as the rule of product. Example:
 * __More Fundamental Counting Principles:__**

In this example, the rule says: multiply 3 by 2, getting 6. [] << resource.

1. The main difference between the two is that in permutations, the order is all important -- we count //abc// as different from //bca//. But in combinations we are concerned only that //a, b,// and //c// have been selected. //abc// and //bca// are the same combination.
 * __Permutations and Combinations__**

Since the order does not matter in combinations, there are going to be fewer combinations then permutations. The combinations are among the permutations- they are a "subset" of the permutations.

For example: Here are all the combinations of //abcd// taken three at a time:  // abc abd acd bcd .//

Each of those 4 combinations will give rise to 3! permutations:


 * abc ||  || abd ||   || acd ||   || bcd ||
 * acb ||  || adb ||   || adc ||   || bdc ||
 * bac ||  || bad ||   || cad ||   || cbd ||
 * bca ||  || bda ||   || cda ||   || cdb ||
 * cab ||  || dab ||   || dac ||   || dbc ||
 * cba ||  || dba ||   || dca ||   || dcb ||

[] << resource  To help you remember, think **p**ermutation...**p**osition, because the position of permutations matter.

In math, combination explains how mnay ways you can arrange something. In math, a combination is an arrangement in which order does not matter. Often contrasted with permutations, which are ordered arrangements, a combination defines how many w ays you could choose a group from a larger group.