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Permutation and combination

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Permutation and combination

  1. 1. DONE BY:SADIA ZAREEN
  2. 2. 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 2H 3H 4H 5H 6H 6*2 = 12 outcomes1T 2T 3T 4T 5T 6T 12 outcomes
  3. 3. 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
  4. 4. 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.
  5. 5. 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 = 6 3! = 3*2*1 = 6ABC ACB BAC BCA CAB CBA
  6. 6. PermutationsTo find the number of Permutations ofn items chosen r at a time, you can usethe formula n! n pr = ( n − r )! where 0 ≤ r ≤ n . 5! 5!5 p3 = = = 5 * 4 * 3 = 60 (5 − 3)! 2!
  7. 7. 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
  8. 8. 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!
  9. 9. 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
  10. 10. 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
  11. 11. 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.
  12. 12. Combinations To find the number of Combinations of n items chosen r at a time, you can use the formula n! C = where 0 ≤ r ≤ n .n r r! ( n − r )!
  13. 13. 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 = = = 10 3 * 2 *1* 2 *1 2 *1 2
  14. 14. 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? Answer Now
  15. 15. 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 = 2,598,960 5* 4* 3* 2*1
  16. 16. 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? Answer Now
  17. 17. 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 * 4 5 C3 = = = = 10 3! (5 − 3)! 3!2! 2 * 1
  18. 18. CombinationsPractice: A basketball team consists of two centers, five forwards, and four guards. In how many ways can the coach select a starting line up of one center, two forwards, and two guards? Answer Now
  19. 19. CombinationsPractice: A basketball team consists of two centers, five forwards, and four guards. In how many ways can the coach select a starting line up of one center, two forwards, and two guards? Center: Forwards: Guards: 2! 5! 5 * 4 4! 4 * 3 2 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.
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