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CHAPTER 2 Counting

CHAPTER 2 Counting. Examples of Counting Problems. Let A ={ a , b , c }. How many subsets of A are there? A penny and a nickel are tossed. How many possible outcomes are there?

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CHAPTER 2 Counting

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  1. CHAPTER 2Counting

  2. Examples of Counting Problems • Let A={a,b,c}. How many subsets of A are there? • A penny and a nickel are tossed. How many possible outcomes are there? • An experiment consists of flipping a coin and then rolling a die. How many different possible outcomes are there? Answers: 8, 4, and 12. These are easy because the sets being counted are small.

  3. Answers • The eight subsets of A are F, {a}, {b},{c},{a,b},{a,c},{b,c},A. • The four possible outcomes are HH, HT, TH, TT. • The twelve possible outcomes are (H,1),(H,2),(H,3),…,(H,6),(T,1),…, (T,6).

  4. A Harder Problem • Of 100 college freshmen, 30 are taking mathematics, 50 are taking English and10 are taking both math and English. The 30 taking math include the 10 taking both and likewise for the 50 taking English. • How many are taking English or math or both? • How many are taking neither?

  5. Solution Let M denote the set of students taking math, let E be the set taking English and let U be the universal set consisting of all 100 students. Refer to the Venn diagram. U E M

  6. Solution(continued)

  7. Addition Principle If A and B are finite sets,

  8. Subtraction Principle(Complement Principle) • If A is finite, • In particular, if U is finite,

  9. Example • How many students are taking English but not math? • Answer:

  10. The Multiplication Principle A local restaurant advertises a fixed price dinner that includes your choice of appetizer, entree, and dessert. If there are 2 appetizers, 3 entrees, and 2 desserts, how many different dinner combinations are there?

  11. D1 D1 D2 D2 E1 E1 A1 D1 D1 E2 E2 D2 D2 D1 D1 E3 E3 D2 D2 A2 Tree Diagram Let A1, A2 denote the appetizers, E1, E2, E3 the entrees, and D1, D2 the desserts. The number of outcomes is the number of final nodes, in this case 12.

  12. The Multiplication Principle(continued) • An “experiment” consists of a sequence of steps S1, S2, …, Sk. • Each Si has ni outcomes, regardless of the outcomes of previous steps. • Then the experiment has outcomes in all.

  13. Example Selecting a Sequence with Replacement • How many sequences of two cards from a standard 52 card deck are there if the second card is allowed to be the same as the first? • Answer: 52×52=2704. • How many sequences of k cards are there if repetitions are allowed? • Answer: 52k.

  14. ExampleSelecting a Sequence without Replacement • How many sequences of 2 cards from a standard deck are there if repetitions are not allowed? • Answer: 52×51=2652 • How many sequences of k cards are there if repetitions are not allowed? • Answer: 52×51×…×(52-k+1) • Note: k≤52

  15. General RuleSelecting with Replacement • How many sequences of two objects from a set of n objects are there if repetitions are allowed? • Answer: n×n • How many sequences of k out of n objects are there if repetitions are allowed? • Answer: n×n×…×n=nk

  16. General RuleSelecting without Replacement • How many sequences of two objects out of a set of n objects are there if repetitions are not allowed? • Answer: n×(n-1) • How many sequences of k out of n objects are there if repetitions are not allowed? • Answer: n×(n-1)×…×(n-k+1) • Note: k≤n

  17. Permutations • Let S be a set of n objects. A permutation of length k from S is a nonrepeating sequence of length k from S. • The number of permutations of length k from a set of n objects is, for k≤n,

  18. The Factorial Function • If k=n, Pn,n=n×(n-1) ×…×2×1 • Also written n!=1×2×…×(n-1)×n • Pronounced “n factorial”. • Examples: 1!=1; 2!=2; 3!=6; 4!=24; 5!=120, 6!=720. • By definition, 0!=1. This makes formulas work out nicely.

  19. Example • King Arthur sits on the North side of the round table. The 12 knights are seated around the table. How many possible arrangements are there? • Answer: 12!=479,001,600. • If Lancelot always sits on Arthur’s left or right, there are 2×11!=79,833,600 possible arrangements (by the multiplication principle.)

  20. Another Formula for Pn,k

  21. Combinations • Let S be a set with n elements. A combination of k elements of S is a subset of S with k elements. • A set is not the same thing as a sequence. {a,b,c} and {c,b,a} are the same set. (a,b,c) and (c,b,a) are different sequences. • Therefore, a combination is not the same as a permutation.

  22. Poker • A five-card draw poker hand is dealt all at once and seen only by the person holding the hand. Therefore, the order of the cards is irrelevant to the betting. A hand is a combination of 5 of the 52 cards. • A five-card stud poker hand is dealt one card at a time and some of the cards are seen by other players. The order of the cards may affect the betting. A hand is a permutation of 5 of the 52 cards.

  23. Counting Combinations • Two-stage experiment resulting in a permutation of k out of n objects: • Stage 1- select a combination of k out of n objects. Let Cn,k denote the number of possible outcomes. • Stage 2- arrange the k selected objects in some order. Pk,k=k! is the number of outcomes.

  24. Counting Combinations(continued) • By the multiplication principle, • We know Pn,kand Pk,k, so • Solving for the unknown,

  25. Draw Poker • How many 5-card draw hands are there in poker? I.e., how many subsets of 5 objects are there in a set of 52 objects? • Answer: • Suggestion: Use the Cn,k button on your calculator. Don’t try to evaluate by hand.

  26. Stud Poker • How many 5-card stud hands are there in poker? I.e., how many permutations of 5 objects from a set of 52 are there? • Answer: P52,5=C52,5×5!=2,598,960×120 =311,875,200.

  27. A Property of Cn,k If S is a set of n elements, then the number of subsets with k elements is the same as the number of subsets with n-k elements because there is a one to one correspondence between sets of size k and their complements of size n-k. I.e., Cn,k=Cn,n-k. This can also be derived algebraically.

  28. Another Example of the Multiplication Principle • A license plate is a sequence of 3 letters followed by a sequence of 3 numerals. Repetitions are allowed. How many license plates are possible? • Answer: There are 263 possible arrangements of the three letters and 103 possible arrangements of the three numerals. Therefore, there are 263× 103=17,576,000 possible license plates.

  29. The Binomial Theorem • The numbers Cn,k occur in the binomial theorem. Sometimes called the binomial coefficients. • Example: • Example: • In general, using summation notation,

  30. How Many Subsets Does a Set Have? • Let S be a set with n elements. • The number of subsets of S is • Expand 2n=(1+1)n by the binomial theorem: • Therefore, S has 2n subsets, including the empty set and S itself.

  31. Example • How many subsets does the set of cards in a standard deck have? • Answer: 252, which is approximately 1016.

  32. Thought Experiment • Your class has n students. You send your principal a coded message telling her which students are going to flunk. The message is a sequence of 0’s and 1’s, where a 1 in position i of the sequence indicates that student number i will flunk. • Use the multiplication principle to determine how many such messages there are. • Comment on the answer.

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