# Counting - PowerPoint PPT Presentation

1 / 27

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:

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

Counting

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

### Counting

• Often in CS we need to

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

• 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

### 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.

### 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.

### 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?

• NEEDED6! / (3! 2!)

• DIVISIBILITY12! / 5!

• AAABBB6! / (3! 3!)

### 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?

### Combinations

• Definition

• Contrast with permutations

• Evaluating

• Applications

• Practice

### 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?

### 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.

### 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.

### 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

### 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?

### 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.

### More counting problems

• Indistinguishable subsets

• Ball-in-urn: the objects being selected are identical

• Pascal’s triangle, a combinatorical argument

• Stirling numbers

### 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.

• 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

### 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).

### 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.

### 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.

### 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.

### 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. 

### 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

### 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!

### 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.

### 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.

### 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.

### 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

### 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)

### 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