# 5.5 counting techniques. more challenging stuff the classical method, when all outcomes are...

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- Slide 1
- 5.5 Counting Techniques
- Slide 2
- More Challenging Stuff The classical method, when all outcomes are equally likely, involves counting the number of ways something can occur This section includes techniques for counting the number of results in a series of choices, under several different scenarios
- Slide 3
- Example If there are 3 different colors of paint (red, blue, green) that can be used to paint 2 different types of toy cars (race car, police car), then how many different toys can there be? 3 colors 2 cars 3 2 = 6 different toys This can be shown in a table or in a tree diagram
- Slide 4
- Table A table of the different possibilities This is a rectangle with 2 rows and 3 columns 2 3 = 6 entries
- Slide 5
- Tree Diagram A tree diagram of the different possibilities Red Blue Green Blue Race Car Blue Police Car Green Race Car Green Police Car Red Race Car Red Police Car Race Police Race Police Race Police Paint Car
- Slide 6
- Multiplication Rule of Counting The Multiplication Rule of Counting applies to this type of situation If a task consists of a sequence of choices With p selections for the first choice With q selections for the second choice With r selections for the third choice Then the number of different tasks is p q r
- Slide 7
- Example Example Part A A child is coloring a picture of a shirt and pants There are 5 different colors of markers How many ways can this be colored? By the multiplication rule 5 5 = 25
- Slide 8
- Different Example Example Part B A child is coloring a picture of a shirt and pants There are 5 different colors of markers The child wants to use 2 different colors How many ways can this be colored? By the multiplication rule 5 4 = 20
- Slide 9
- Example continued Allowing the same marker to be used twice 5 5 = 25 Requiring that there be two different markers 5 4 = 20 There are 5 selections for the first choice for both Part A and Part B of this example But they differ for the second choice there are only 4 selections for Part B
- Slide 10
- Repetition?? Example continued Part A, allowing the same marker to be used twice, is called counting with repetition and has formulas such as 5 5 5 Part B, requiring that there be two different markers, is called counting without repetition and has formulas such as 5 4 3
- Slide 11
- Calculator Commands While in a CALCULATOR Page: Menu, Probability, We will be using Factorial, permutations, & combinations
- Slide 12
- Factorial One way to help write these products is using the factorial symbol n! n! = n (n-1) (n-2) 2 1 We start off by saying that 0! = 1 and 1! = 1 For example 5! = 5 4 3 2 1 = 120 Notice how 5! looks like the 5 4 3 from the previous example
- Slide 13
- Permutation (Order Matters) The problem of choosing one marker out of 5 and then choosing a second marker out of the 4 remaining is an example of a permutation A permutation is an ordered arrangement, in which r different objects are chosen out of n different objects with repetition not allowed The number of ways is written n P r
- Slide 14
- Permutation Formula A mathematical way to write the formula for the number of permutations is This is a very convenient mathematical way to write a formula for n P r, but it is not a particularly efficient way to actually compute it In particular, n! gets rapidly gets very large
- Slide 15
- Order For some problems, the order of choice does not matter Order matters example Choosing one person to be the president of a club and another to be the vice-president Two different roles Order does not matter example Choosing two people to go to a meeting The same role
- Slide 16
- Combination (Order Does Not Matter) When order does not matter, this is called a combination A combination is an unordered arrangement, in which r different objects are chosen out of n different objects with repetition not allowed The number of ways is written n C r
- Slide 17
- Permutation vs. Combination Comparing the description of a permutation with the description of a combination The only difference is whether order matters
- Slide 18
- Combination Formula Because each combination corresponds to r! permutations, the formula n C r for the number of combinations is
- Slide 19
- Example If there are 8 researchers and 3 of them are to be chosen to go to a meeting A combination since order does not matter There are 56 different ways that this can be done
- Slide 20
- Permutation or Combination Is a problem a permutation or a combination? One way to tell Write down one possible solution (i.e. Roger, Rick, Randy) Switch the order of two of the elements (i.e. Rick, Roger, Randy) Is this the same result? If no this is a permutation order matters If yes this is a combination order does not matter
- Slide 21
- Different Our permutation and combination problems so far assume that all n total items are different Sometimes we have a permutations but not all of the n items are different This is a more complicated problem How many ways are there?
- Slide 22
- Example How many ways to put 3 As, 2 Ns, and 2 Ts to try to make a seven letter sequence? ____ ____ ____ ____ ____ ____ ____ Each of the blanks can be filled in with either an A or a N or a T The three As are the same the two Ns are the same the two Ts are the same
- Slide 23
- Example Continued Where can the As go? There are 7 possible places Any 3 of them are possible Order does not matter So 7 C 3 different ways to put in the As
- Slide 24
- Example Continued Where can the Ns go? There are 4 possible places (since 3 of the 7 have been taken by the As already) Any 2 of them are possible Order does not matter So 4 C 2 different ways to put in the Ns And there are 2 C 2 different ways to put in the Ts
- Slide 25
- Example Continued Altogether there are 7 C 3 4 C 2 2 C 2 different ways This is Notice that the denominator is 3, 2, 2 the numbers of each letter
- Slide 26
- Permutation A permutation example In a horse racing Trifecta, a gambler must pick which horse comes in first, which second, and which third If there are 8 horses in the race, and every order of finish is equally likely, what is the chance that any ticket is a winning ticket? Order matters, so this is a permutations problem
- Slide 27
- Permutation Cont. A permutation example continued There are 8 P 3 permutations of the order of finish of the horses The probability that any one ticket is a winning ticket is 1 out of 8 P 3, or 1 out of 336
- Slide 28
- Combination Example A combination example The Powerball lottery consists of choosing 5 numbers out of 55 and then 1 number out of 42 The grand prize is given out when all 6 numbers are correct What is the chance of getting the grand prize? Order does not matter, so this is a combinations problem (for the 5 balls)
- Slide 29
- Combination Cont. A combination example continued There are 55 C 5 combinations of the 5 numbers There are 42 possibilities for the last ball, so the probability of the grand prize is 1 out of which is pretty small
- Slide 30
- Summary The Multiplication Rule counts the number of possible sequences of items Permutations and combinations count the number of ways of arranging items, with permutations when the order matters and combinations when the order does not matter Permutations and combinations are used to compute probabilities in the classical method