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- 1. Permutations andCombinations AII.12 2009

2. Objectives: apply fundamental counting principle compute permutations compute combinations distinguish permutations vs combinations 3. Fundamental Counting PrincipleFundamental Counting Principle can be useddetermine the number of possible outcomeswhen there are two or more characteristics .Fundamental Counting Principle states thatif an event has m possible outcomes andanother independent event has n possibleoutcomes, then there are m* n possibleoutcomes for the two events together. 4. Fundamental Counting PrincipleLets start with a simple example. A student is to roll a die and flip a coin.How many possible outcomes will there be?1H 2H3H 4H 5H 6H 6*2 = 12 outcomes1T 2T3T 4T 5T 6T12 outcomes 5. Fundamental Counting PrincipleFor a college interview, Robert has to choosewhat to wear from the following: 4 slacks, 3shirts, 2 shoes and 5 ties. How many possibleoutfits does he have to choose from?4*3*2*5 = 120 outfits 6. Permutations A Permutation is an arrangement of items in a particular order. Notice, ORDER MATTERS!To find the number of Permutations ofn items, we can use the FundamentalCounting Principle or factorial notation. 7. Permutations The number of ways to arrange the letters ABC:____ ________Number of choices for first blank? 3 ____ ____Number of choices for second blank?3 2 ___Number of choices for third blank? 3 2 1 3*2*1 = 63! = 3*2*1 = 6ABC ACB BACBCA CAB CBA 8. PermutationsTo find the number of Permutations ofn items chosen r at a time, you can usethe formulan! n pr = ( n r )! where 0 r n .5!5!5 p3 = = = 5 * 4 * 3 = 60 (5 3)! 2! 9. PermutationsPractice: A combination lock will open when the right choice of three numbers (from 1 to 30, inclusive) is selected. How many different lock combinations are possible assuming no number is repeated?Answer Now 10. PermutationsPractice: A combination lock will open when the right choice of three numbers (from 1 to 30, inclusive) is selected. How many different lock combinations are possible assuming no number is repeated? 30!30!30 p3 = = = 30 * 29 * 28 = 24360( 30 3)! 27! 11. PermutationsPractice: From a club of 24 members, a President, Vice President, Secretary, Treasurer and Historian are to be elected. In how many ways can the offices be filled? Answer Now 12. PermutationsPractice: From a club of 24 members, a President, Vice President, Secretary, Treasurer and Historian are to be elected. In how many ways can the offices be filled? 24!24!24 p5 = = =( 24 5)! 19!24 * 23 * 22 * 21 * 20 = 5,100,480 13. Combinations A Combination is an arrangement of items in which order does not matter.ORDER DOES NOT MATTER!Since the order does not matter incombinations, there are fewer combinationsthan permutations. The combinations are a"subset" of the permutations. 14. Combinations To find the number of Combinations of n items chosen r at a time, you can use the formulan! C =where 0 r n .n r r! ( n r )! 15. CombinationsTo find the number of Combinations ofn items chosen r at a time, you can usethe formula n!C =where 0 r n . n r r! ( n r )! 5! 5!5 C3 = = =3! (5 3)! 3!2!5 * 4 * 3 * 2 * 1 5 * 4 20 = = = 103 * 2 *1* 2 *1 2 *1 2 16. CombinationsPractice:To play a particular card game, eachplayer is dealt five cards from astandard deck of 52 cards. Howmany different hands are possible?Answer Now 17. CombinationsPractice: To play a particular card game, each player is dealt five cards from a standard deck of 52 cards. How many different hands are possible?52!52! 52 C5 = == 5! (52 5)! 5!47! 52 * 51 * 50 * 49 * 48= 2,598,9605* 4* 3* 2*1 18. CombinationsPractice:A student must answer 3 out of 5essay questions on a test. In howmany different ways can thestudent select the questions?Answer Now 19. CombinationsPractice: A student must answer 3 out of 5 essay questions on a test. In how many different ways can the student select the questions?5!5! 5 * 45 C3 ==== 10 3! (5 3)! 3!2! 2 * 1 20. CombinationsPractice:A basketball team consists of twocenters, five forwards, and fourguards. In how many ways can thecoach select a starting line up ofone center, two forwards, and twoguards?Answer Now 21. CombinationsPractice: A basketball team consists of two centers, five forwards,and four guards. In how many ways can the coach select astarting line up of one center, two forwards, and twoguards?Center:Forwards:Guards:2! 5! 5 * 4 4! 4 * 32 C1 ==2 5 C2 = == 10 4 C2 = ==6 1!1! 2!3! 2 * 1 2!2! 2 * 1 2 C1 * 5 C 2 * 4 C 2 Thus, the number of ways to select the starting line up is 2*10*6 = 120.