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# Counting

Counting. Often in CS we need to Enumerate/account for all possibilities of something, such as: generating passwords exhaustive testing of an algorithm Calculate a probability In general, there are two ways to count:

## Counting

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### Presentation Transcript

1. Counting • Often in CS we need to • Enumerate/account for all possibilities of something, such as: • generating passwords • exhaustive testing of an algorithm • Calculate a probability • In general, there are two ways to count: • Chicken method  We’ll get an answer eventually, but we won’t necessarily understand why • Analytically  but we need to learn some rules

2. Counting outline • Some types of problems to address • Definition of probability • Based on some “experiment” or “event” that has a set of outcomes called our sample space. • P = (# favorable outcomes)/ (total in sample space) • Multiplication rule, i.e. fundamental counting principle • Number of possible phone numbers or license plate numbers • Permutations • Combinations • *** See handout for details • Common technique required: break a problem down into distinct cases.

3. Multiplication rule • If you need to make consecutive (or independent) choices, just multiply. • If you have x ways of making one decision, and y ways of making a second decision, the total number of outcomes is xy. • We can generalize to a “lining up” situation. If you have x items that need to be arranged in some order, this can be done in x! ways. • Sometimes we have different cases to add. • E.g. At 3:00, you may choose from either the lunch or dinner menu. Do lunch and dinner separately, then add.

4. Distinguishable • The number of ways to line up n objects, but a of them are identical, b of them are identical, etc. = n! / (a! b! …) • Think of 3 people sitting in a row of 3 chairs versus 5 chairs. • Examples: How many ways can we arrange the letters in the following words? • NEEDED 6! / (3! 2!) • DIVISIBILITY 12! / 5! • AAABBB 6! / (3! 3!)

5. Handling restriction • (This is a permutation problem with a twist.) • We need to select a president, vice president and secretary from 10 people. However, A can’t be president and B cannot be secretary. How many ways can this be done? • Focus on cases that are legal: • B is president, A is VP, anyone else can by sec’y • B is president, A is secretary, anyone else can be VP • B is VP, A is sec’y, anyone else can be president • Any more?

6. Combinations • Definition • Contrast with permutations • Evaluating • Applications • Practice

7. When to use • Questions dealing with making some selection, membership, inclusion, presence vs. absence…. In which order does not matter • Combinations answer this type of question: A set has n elements. How many subsets contain r elements? • Try examples when n = 4. • Contrast with permutations: • How many ways can we select 3 letters from the alphabet? • How many 3 letter words can we make?

8. Points to remember • Order doesn’t matter • Subsets, committees • What matters is whether something is chosen or not. • Evaluation: • The number of ways to choose r items out of a set of n is n! / (r!(n – r)!). • Shorthand notation: C(n, r) or • Shortcut to evaluate: C(10, 3) = 10 * 9 * 8 / (3 * 2 * 1) • Incidentally: choosing r is the same as (not) choosing n – r.

9. Examples • How many 8-bit representations have exactly three 0’s? • Choose which positions get a 0: choose 3 out of 8: C(8, 3). • Six people need to be divided into two teams A and B. How many ways can this be done? • Choose 3 to go to team A: C(6, 3). • Choose 3 remaining people for B: C(3, 3) = 1. • What is the probability that exactly 4 out of 7 coin tosses will be heads? • Choose which 4 occurrences are heads: C(7, 4). • Divide by the total size of sample space: 27.

10. Watch the wording • Sometimes we want “at least” or “at most” • Both • Neither • Be careful with “or”. There could be overlap. • King or club • Number or red • Are objects/categories distinguishable? • “two teams” may be indistinguishable • Selecting cards from a deck: almost always the order that we deal cards does not matter

11. Indirect probability • Sometimes it’s easier to find the probability that an even will NOT happen, and then subtract from 1. • Ex. Suppose we roll 5 dice, as in Yahtzee. • P(all sixes) = 1/7776 • P(didn’t get all sixes) = 1 – 1/7776 • P(1000 tries, didn’t get all sixes) = (1 – 1/7776)1000 • P(1000 tries, got all sixes at least once) = 1 – (1 – 1/7776)1000 • Ex. The “birthday problem.” In a group of n people, what is the probability that at least two of them have the same birthday? What is n if p > ½?

12. Poker hands • How many straight flushes are there? • The high card can be anything from 5 up to Ace. • All cards of the same suit, so select a suit. • How many 4-of-a-kind hands? • Choose which denomination you want 4 of. • Choose those 4 cards. • Choose the 5th card. • Full house? • Choose two denominations. • Choose three of one and two of the other.

13. More counting problems • Indistinguishable subsets • Ball-in-urn: the objects being selected are identical • Pascal’s triangle, a combinatorical argument • Stirling numbers

14. Indistinguishable subsets • How many ways can 4 people divide into 2 teams of 2, as in a doubles match? • The teams are the same size and don’t have inherent name, place, uniform, etc. • AB/CD, AC/BD, AD/BC • Is the answer 3 or 6? • How many ways to divide … ? • 20 people into 4 teams of 5 • 20 people into teams of 3, 4, 5, 8 • 20 people into teams of 4, 5, 5, 6

15. Ball in urn • The number of ways to distribute • n distinct objects into r categories = rn • n identical objects into r categories = C(n + r – 1, n) • Give 4 cards to 2 people. • If face up, they are distinct. Each card has 2 possible destinations  24. • If face down, then the only thing that matters is how many cards you have. Your cards and my cards are separated by a divider CARD CARDCARD / CARD 0 0 0 1 0 2 categories  1 divider that can go anywhere: answer = C(5, 1) or C(5, 4).

16. Ball in urn (2) • Give 5 cards to 3 people. • If distinct, ask each card where it’s going: 35. • If identical: CARD CARD / CARD CARD / CARD 0 0 1 0 0 1 0 3 categories  2 dividers that can go anywhere Total = C(7, 2) • 3 divers find 5 gold coins. How many ways to share? • A case of 12 bottles of wine to distribute among 4 tables.

17. Ball in urn (3) • What if we must give each table at least 1 bottle of wine? • We have less freedom to choose. • Only 8 bottles can be freely distributed. • Let’s buy 7 cans of soup. 3 varieties are available.

18. Ride tickets • Buy 10 ride tickets at the fair. There are 3 possible rides. • If the issue is how many times you want to go on each, this is a ball in urn problem  C(10 + 3 – 1, 2) • If order matters  310 • If the tickets say we must go in haunted house 4 times, roller coaster 3 times and ferris wheel 3 times, and order matters. It’s a combination question. Choose when you do each type of ride  C(10, 4) * C(6, 3) * C(3, 3). • If the tickets are bought 4x, 3x and 3x, and order doesn’t matter  1. • *** More examples from book and handout.

19. Understanding C(n, r) • Let’s justify: C(5, 2) = C(4, 1) + C(4, 2) • We want to select two elements from { A, B, C, D, E }. • Is E part of the selection? • If so, then we pick one of the others: C(4, 1) • If not, then we pick both from the others: C(4, 2). • This line of reasoning works in general for C(n, r). • Now that we understand Pascal’s triangle, let’s look at another one. 

20. What good is Pascal’s Δ • Combinatorical arguments • C(n, r)  numbers in Pascal’s triangle  • Stirling numbers • Recap important ideas • When to use formulas: n!, P(n, r), C(n, r), C(n + r – 1, n) • Go over practice handout

21. Stirling numbers • S(n, r) = the number of ways to break up a set of n elements into r groups. • If you study these numbers in detail, you can solve more complex counting problems, and determine coefficients of more Bernoulli-type summations!

22. Starting simple • Let’s consider S(3, 2). How many ways can we split up a 3-element set { A, B, C } into two groups? Note: their sizes must be 1 and 2. • A / BC • B / AC • C / AB • Consider S(3, 3). Splitting { A, B, C } into 3 groups. • A / B / C • Next, let’s work out various cases of S(5, r) for 1  r  5.

23. S(5, r) • S(5, 1): We need to “split up” a set of 5 elements into a single group. Degenerate case: we want 1 group of size 5. Answer = 1. • S(5, 2): Split into 2 groups • Their sizes could be 2 and 3. In this case, we have groups like A B / C D E. Select which 2 go in one group: C(5, 2). • Their sizes could be 1 and 4. In this case, we have groups like A / B C D E. Select 1 to go into the first group: C(5, 1). • Answer is the sum: 10 + 5 = 15.

24. S(5, r) continued • S(5, 3): We need to split into 3 groups • The sizes could be 1, 1, 3. For example A / B / CDE. Think of it in terms of distinguishable permutations. The 5 elements can be ordered in 5! ways. But the 3 elements in the size-3 group have no inherent order, so we have to divide by 3!. Also, the two groups of size 1 are indistinguishable, so we divide by another 2!. 5!/(3!2!) = 10 • The sizes could be 1, 2, 2. For example A / BC / DE. The total number of permutations is 5!. But the 2 groups of 2 are indistinguishable. Also, the pairs of values inside each group of 2 are indistinguishable. 5!/(2!2!2!) = 15. • Total = 25.

25. S(5, r) continued • S(5, 4): We need to split into 4 groups • The sizes could only be 1, 1, 1, 2. The two positions inside the size-2 group are not distinguishable, and the 3 groups of size 1 are not distinguishable. 5!/(2!3!) = 10. • S(5, 5): pulverize the set of 5 into singleton sets. Just one way to do it. (You can think of all 5 size-1 groups as being indistinguishable, and 5!/5! = 1) • So, row 5 of Stirling’s triangle is: 1, 15, 25, 10, 1

26. General approach • It’s a bit tedious to keep evaluating each S(n, r) from scratch in terms of every possible case of group size. • Let’s make use of earlier/lower values of S. • Consider S(6, 3). Split { A, B, C, D, E, F } into 3 parts. • When we partition into groups, we may have F by itself. • F / (2 more groups of A-E). Answer = S(5, 2). • After partitioning A-E, just put F into any current group. There are 3 to choose from. Answer = 3 * S(5, 3) • Add it up: S(5, 2) + 3 * S(5, 3) • In general: S(n, r) = S(n – 1, r – 1) + r * S(n – 1, r)

27. Using the formula • How can we use the Stirling formula to quickly determine S(n, r) for higher values of n? • Cell value = NW + N * (column number) • Observe 90 = 15 + 25*3 and 65 = 25 + 10*4

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