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Counting

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- 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

- Enumerate/account for all possibilities of something, such as:
- 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

- 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.

- 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.

- 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?
- NEEDED6! / (3! 2!)
- DIVISIBILITY12! / 5!
- AAABBB6! / (3! 3!)

- (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?

- Definition
- Contrast with permutations
- Evaluating
- Applications
- Practice

- 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?

- 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.

- 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.

- 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

- 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?

- 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.

- Indistinguishable subsets
- Ball-in-urn: the objects being selected are identical
- Pascal’s triangle, a combinatorical argument
- Stirling numbers

- 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

- 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).

- 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.

- 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.

- 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.

- 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.

- 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

- 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!

- 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.

- 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.

- 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.

- 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

- 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)

- 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