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CS1022 Computer Programming & Principles

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CS1022Computer Programming & Principles

Lecture 6.1

Combinatorics (1)

- Motivation
- Addition principle
- Multiplication principle
- Analysing selection problems
- k-Samples
- k-Permutations
- k-Combinations
- k-Selections
- Examples

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- We sometimes need to consider the question
- How many elements are there in set A?

- If too many elements in a set, and we need to process the set, then this might take too long
- What’s “too long”?
- Example: how many outfits can one wear?
- Blue trousers and yellow shirt
- Black trousers and yellow shirt
- Etc.

- When we are proposing a computational solution we need to think of feasible solutions
- In realistic situations, will the program finish in “good time”?

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- Every term we organise your timetable
- Day/time for lectures, practicals, etc.
- Compulsory modules must be allowed
- Not all choices possible, though

- How many ways are there to
- Arrange lecture, practical and lab times
- For all modules, for the whole university,
- Using up all time slots, 9AM-5PM?

- Answer: many!
- A program to compute a timetable needs to consider every option
- It is going to take a long time...
- International timetabling competition!

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- Disjoint choices/events don’t influence one another
- Choice of cakes in a shop – first choice influences second one as there will be fewer cakes
- Throwing a die twice – 2nd time not influenced by 1st time

- Suppose A and Bdisjoint choices/events
- The number of options/outcomes for A is n1
- The number of options/outcomes for B is n2

- Total number of options/outcomes for A or B is
n1 n2

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- Suppose a sequence of k choices/events
- With n1 options/outcomes for 1st choice/event
- With n2 options/outcomes for 2nd choice/event
- ...
- With nk options/outcomes for kth choice/event

- Total number of possible options/outcomes for the sequence is
n1 n2 ... nk

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- Suppose A is the set of n1 options/outcomes
- Suppose B is the set of n2options/outcomes
- Sets A and B are disjoint, that is, A B
- So |A B| |A| |B|
- That is, A B contains n1 n2 elements
- There are n1 n2 possible outcomes for AorB

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- We can “interpret” multiplication principle in terms of sets too
- Let A1 be set of n1 options/outcomes for 1st choice/event
- Let A2 be setof n2 options/outcomesfor 2nd choice/event
- ...
- Let Ak be set of nk options/outcomes for kth choice/event

- Sequence of k events is an element of Cartesian product A1 A2 ... Ak
- Set has cardinality |A1| |A2| ...|Ak|

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- Suppose we need to create a way to identify cars
- What kind of licence plate would you propose?

- Some ways are better than others
- How many cars could we licence with a 6-digit licence?
- Answer: 1,000,000, that is from 000000 to 999999
- UK: 40,000,000 vehicles so we need more digits...

- Alternative – licence with 4 letters and 3 digits
- Examples: ABCD 123, GHHH 234, etc.
- How many cars can we licence?
- Answer: 26 26 26 26 10 10 10 = 456,976,000

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- Suppose we are offered 3 kinds of sweets
- Aniseed drops (A)
- Butter mints (B) and
- Cherry drops (C)

- How many ways can we choose 2 sweets?
- Can we choose the same sweet twice? (AA, BB or CC?)
- Does the order matter? (Is BA the same as AB?)

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- Cases to consider
- Repeats allowed and order matters
- 9 possibilities: AA, AB, AC, BA, BB, BC, CA, CB, CC

- Repeats allowed and order does not matter
- 6 possibilities: AA, AB, AC, BB, BC, CC

- Repeats not allowed and order matters
- 6 possibilities: AB, AC, BA, BC, CA, CB

- Repeats not allowed and order does not matter
- 3 possibilities: AB, AC, BC

- Repeats allowed and order matters

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- Consider
- Selection of k objects from a set of n objects where
- Order mattersand repetition is allowed

- Any such selection is called a k-sample
- Since repetition is allowed,
- There are n ways of choosing the 1st object, and
- There are n ways of choosing the 2nd object, and so on
- Until all k objects have been selected

- By the multiplication principle this gives
n n ... n = nkpossible k-samples

k

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- Consider
- Selection of k objects from a set of n objects where
- Order mattersand repetition is not allowed

- Any such selection is called a k-permutation
- Total number of k-permutations is denoted by P(n, k)

- Since repetition is not allowed,
- There are n ways of choosing the 1st object
- There are (n – 1) ways of choosing the 2nd object
- There are (n – 2) ways of choosing the 3rd object, and so on
- Up to (n – k 1) ways of choosing the kth object

- By the multiplication principle this gives
P(n, k) n(n – 1)(n – k 1) n!/ (n – k)!

possible k-permutations

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- How many 4-letter “words” can we write with distinct letters from the set a, g, m, o, p, r?
- Solution:
- A “word” is any ordered selection of 4 different letters
P(n, k) n!

(n – k)!

We have in our scenario n 6 and k 4, so we have

P(6, 4) 6!/ (6 – 4)! 6!/2!

(6 5 4 3 2 1)/(2 1)

(6 5 4 3 2 1)/(2 1) 360

possible k-permutations

- A “word” is any ordered selection of 4 different letters

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- Consider
- Selection of k objects from a set of n objects where
- Order does not matterand repetition is not allowed

- Any such selection is called a k-combination
- Total number of k-combinations is denoted by C(n, k)

- By the multiplication principle:
- The number of permutations of k distinct objects selected from n objects is
- The number of unordered ways to select the objects multiplied by the number of ways to order them

- Hence, P(n, k) C(n, k) k! and so there are
C(n, k) P(n, k) n!

k! (n – k)! k!

possible k-combinations

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- Consider
- Selection of k objects from a set of n objects where
- Order does not matterand repetition is allowed

- Any such selection is called a k-selection
- Since selection is unordered,
- We arrange the k objects so that like objects grouped together and separate the groups with markers
- There are n ways of choosing the 2nd object, and so on

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- Example: unordered selection of 5 letters from collection a, b and c, with repetitions
- Selection of two a’s, one b and two c’s: aa|b|cc
- Selection of one a’s, and four c’s: a||cccc
- Seven slots: five letters and two markers
- Different choices: ways to insert 2 markers into 7 slots
- That is, C(7, 2)

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- Unordered selection of k objects from a set of n objects with repetition allowed requires
(n – 1) markers and k objects

- Therefore, there are (n – 1) k slots to fill
- The number of k-selections is the number of ways of putting (n – 1) markers into the (n – 1) k slots

- Therefore number of k-selections from n objects is
C((n k – 1), (n – 1)) (n k – 1)!

((n k – 1) – (n – 1))! (n – 1)!

(n k – 1)!

k! (n – 1)!

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- Five dice are thrown. How many different outcomes are possible?
- Solution:
- Each of the dice shows one of six outcomes
- If 5 dice are thrown, the number of outcomes is the unordered selection of 5 objects with repetition allowed
- C((n k – 1), (n – 1)), with n 6 and k 5
- This gives
C((6 5– 1), (6 – 1)) C(10, 5) 10!/(5! 5!) 252

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- What we have so far can be summarised as
- We need to analyse the problem to choose formula
- They all need as input n and k
- If we choose the wrong formula, we still get a result...
- But the result will not be the right answer!

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- Twice-weekly draw selects 6 different numbers
- Random selection from {1, 2, ..., 49}
- People pick their numbers before draw (and pay for this)

- Let’s calculate the odds of hitting the jackpot
- Winning numbers: one of the unordered selection of six numbers from 49 possibilities
- That is, C(49, 6) = 13,983,816 different ways
- The odds are 1 to 13,983,816

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- Smaller prizes for getting 5, 4 or 3 numbers right
- If you choose exactly 3 numbers right you get £10

- What are the odds of getting £10?
- Pre-selection of numbers means choosing three correct and three incorrect numbers
- There are C(6,3) ways of selecting 3 correct numbers
- There are C(43,3) ways of selecting 3 incorrect numbers
- Total number of winning combinations is
- Odds of 13,983,816/246,820 57 to 1

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- 12 candidates for a committee of 5 people
- How many committees?
- Order does not matter
- Elements not repeated

- There are
possible committees

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- We want to know about two people, Mary & Peter
- How many committees contain both Mary & Peter?
- Reasoning
- If Mary & Peter are in the committee, then we need only consider the remaining 3 people to complete the group
- There are 10 people to fill in the 3 places
- Order does not matter
- Elements not repeated

- There are
possible committees with Mary and Peter in them

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- How many do not contain neither Mary or Peter?
- Reasoning
- If Mary & Peter are excluded, then we have to select 5 people from remaining 10 people
- Order does not matter
- Elements not repeated

- There are
possible committees without Mary and Peter

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- R. Haggarty. “Discrete Mathematics for Computing”. Pearson Education Ltd. 2002. (Chapter 6)
- Wikipedia’s entry
- Wikibooks entry

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