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- Slide 1
- COUNTING PRINCIPALS, PERMUTATIONS, AND COMBINATIONS
- Slide 2
- A COUNTING PROBLEM ASKS HOW MANY WAYS SOME EVENT CAN OCCUR. Ex. 1: How many three-letter codes are there using letters A, B, C, and D if no letter can be repeated? One way to solve is to list all possibilities. ABCABD ACBACDADBADC BAC BADBCABCDBDABDC CABCADCBACBDCDACDB DABDACDBADBCDCADCB
- Slide 3
- Ex. 2: An experimental psychologist uses a sequence of two food rewards in an experiment regarding animal behavior. These two rewards are of three different varieties. How many different sequences of rewards are there if each variety can be used only once in each sequence? Next slide
- Slide 4
- Another way to solve is a factor tree where the number of end branches is your answer. Count the number of branches to determine the number of ways. There are six different ways to give the rewards.
- Slide 5
- FUNDAMENTAL COUNTING PRINCIPLE Suppose that a certain procedure P can be broken into n successive ordered stages, S 1, S 2,... S n, and suppose that S 1 can occur in r 1 ways. S 2 can occur in r 2 ways. S n can occur in r n ways. Then the number of ways P can occur is
- Slide 6
- Ex. 2 Revisited: An experimental psychologist uses a sequence of two food rewards in an experiment regarding animal behavior. These two rewards are of three different varieties. How many different sequences of rewards are there if each variety can be used only once in each sequence? Using the fundamental counting principle: 3 2 X 1st reward 2nd reward
- Slide 7
- PERMUTATIONS An r-permutation of a set of n elements is an ordered selection of r elements from the set of n elements ! means factorial Ex. 3! = 321 0! = 1
- Slide 8
- Ex. 1 Revisited: How many three-letter codes are there using letters A, B, C, and D if no letter can be repeated? Note: The order does matter
- Slide 9
- COMBINATIONS The number of combinations of n elements taken r at a time is Where n & r are nonnegative integers & r < n Order does NOT matter!
- Slide 10
- Ex. 3: How many committees of three can be selected from four people? Use A, B, C, and D to represent the people Note: Does the order matter?
- Slide 11
- Ex. 4: How many ways can the 4 call letters of a radio station be arranged if the first letter must be W or K and no letters repeat?
- Slide 12
- Ex. 5: In how many ways can a class of 25 students elect a president, vice- president, and secretary if no student can hold more than one office?
- Slide 13
- Ex. 6: How many five-card hands are possible from a standard deck of cards?
- Slide 14
- Ex. 7: In how many ways can 9 horses place 1 st, 2 nd, or 3 rd in a race?
- Slide 15
- Ex. 8: Suppose there are 15 girls and 18 boys in a class. In how many ways can 2 girls and 2 boys be selected for a group project? 15 C 2 18 C 2 = 16,065
- Slide 16
- ITS TIME TO COMBINE & EXPAND From a group of 7 men and 6 women, five people are to be selected to form a committee so that at least 3 men on the committee. In how many ways can it be done? What are the possible scenarios 3 men and 2 women, 4 men and 1 woman, or 5 men ( 7 C 3 x 6 C 2 ) + ( 7 C 4 x 6 C 1 ) + ( 7 C 5 ) - but why do we add here? (525 + 210 + 21) 756
- Slide 17
- MORE EXAMPLES In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together? The word 'LEADING' has 7 different letters. But, when we pair all the vowels together (EAI) they count as 1 letter So there are 5 letters 5! = 120 combinations But the vowels can also be rearranged, so 3! = 6 When we combine these two we get 120 x 6 = 720 possible solutions
- Slide 18
- CONTINUED In how many different ways can the letters of the word 'DETAIL' be arranged in such a way that the vowels occupy only the odd positions? There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants. Let us mark these positions as under: (1) (2) (3) (4) (5) (6) Now, 3 vowels can be placed at any of the three places marked 1, 3, 5. Number of ways of arranging the vowels = 3 P 3 = 3! = 6. Also, the 3 consonants can be arranged at the remaining 3 positions. Number of ways of these arrangements = 3 P 3 = 3! = 6. Total number of ways = (6 x 6) = 36.