Permutation and combination
- 2. Fundamental Counting Principle Lets 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 outcomes 1T 2T 3T 4T 5T 6T 12 outcomes
- 3. Fundamental Counting Principle For a college interview, Robert has to choose what to wear from the following: 4 slacks, 3 shirts, 2 shoes and 5 ties. How many possible outfits does he have to choose from? 4*3*2*5 = 120 outfits
- 4. Permutations A Permutation is an arrangement of items in a particular order. Notice, ORDER MATTERS! To find the number of Permutations of n items, we can use the Fundamental Counting Principle or factorial notation.
- 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 = 6 ABC ACB BAC BCA CAB CBA
- 6. Permutations To find the number of Permutations of n items chosen r at a time, you can use the formula n! n pr = ( n − r )! where 0 ≤ r ≤ n . 5! 5! 5 p3 = = = 5 * 4 * 3 = 60 (5 − 3)! 2!
- 7. Permutations Practice: 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. Permutations Practice: 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. Permutations Practice: 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. Permutations Practice: 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. Combinations A Combination is an arrangement of items in which order does not matter. ORDER DOES NOT MATTER! Since the order does not matter in combinations, there are fewer combinations than permutations. The combinations are a "subset" of the permutations.
- 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. 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 )! 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. Combinations Practice: 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. Combinations Practice: 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. Combinations Practice: 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. Combinations Practice: 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. Combinations Practice: 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. Combinations Practice: 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.