1. CS1022 Computer Programming & Principles
Lecture 6.1
Combinatorics (1)

2. Plan of lecture Motivation Addition principle Multiplication principle
Analysing selection problems
k-Samples
k-Permutations
k-Combinations
k-Selections
Examples
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3. Why combinatorics? 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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4. A scenario: timetabling
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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5. Addition principle 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 B disjoint 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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6. Multiplication principle
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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7. Addition principle with sets
Suppose A is the set of n1 options/outcomes
Suppose B is the set of n2 options/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 A or B
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8. Multiplication principle with sets
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 set of n2 options/outcomes for 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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9. Analysing information needs
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 to
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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10. Analysing selection problems (1)
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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11. Analysing selection problems (2)
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
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12. n  n  ...  n = nk possible k-samples
Consider
Selection of k objects from a set of n objects where
Order matters and 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 = nk possible k-samples k
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13. P(n, k)  n(n – 1)(n – k  1)  n!/ (n – k)!
k-Permutations (1)
Consider
Selection of k objects from a set of n objects where
Order matters and 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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14. k-Permutations (2)
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
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15. k-Combinations (1) Consider
Selection of k objects from a set of n objects where
Order does not matter and 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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16. k-Selection (1) Consider Any such selection is called a k-selection
Selection of k objects from a set of n objects where
Order does not matter and 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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17. k-Selection (2)
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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18. (n – 1) markers and k objects
k-Selection (3)
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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19. k-Selection (4)
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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20. Summarising... 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!
Order matters
Order does not matter
Elements repeated
k-sample: nk
k-selection:
Elements not repeated
k-permutation:
k-combination:
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21. Example: National Lottery (1)
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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22. Example: National Lottery (2)
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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23. Example: Choosing people (1)
12 candidates for a committee of 5 people
How many committees?
Order does not matter
Elements not repeated
There are
possible committees
Order matters
Order does not matter
Elements repeated nk
Elements not repeated
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24. Example: Choosing people (2)
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
Order matters
Order does not matter
Elements repeated nk
Elements not repeated
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25. Example: Choosing people (3)
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
Order matters
Order does not matter
Elements repeated nk
Elements not repeated
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26. Further reading
R. Haggarty. “Discrete Mathematics for Computing”. Pearson Education Ltd (Chapter 6)
Wikipedia’s entry
Wikibooks entry
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