COMP3010: Algorithm Theory and Design

Daniel Sutantyo,  Department of Computing, Macquarie University

3.1 Permutations and Combinations

Search space

3.1 - Permutations and Combinations

  • The set of all possible solution is known as the search space or solution space
  • When we are brute forcing, we simply go through all possible solution and pick the correct one
  • Obviously, we need to be able to generate all possible solutions, and to do this we need to employ techniques from combinatorial mathematics (which for us is just permutations and combinations)
    • references: CLRS, Appendix C, Section C.1

 

Rule of Sum

  • The number of ways to choose one element from one or more disjoint sets is the sum of the cardinalities of the sets
    • example: a character on the number plate can either be a number of a letter, how many choices do you have?
      • set of 1-digit numbers: \(\{0,1,2,3,4,5,6,7,8,9\}\), cardinality is 10
      • set of letters: \(\{A,B,C,\dots,Y,Z\}\), cardinality is 26
      • number of choices: 26 + 10 = 36 choices

3.1 - Permutations and Combinations

Rule of Product

  • The number of ways to choose an ordered pair is the number of ways of choosing the first element multiplied by the number of ways of choosing the second element
    • example: a single character followed by a one-digit integer (e.g. \(A9, C5, Z3\)), has \(26 * 10 = 260\) combinations
    • example: there are 100 2-digit numbers (from 00 to 99)

3.1 - Permutations and Combinations

  • A permutation of a finite set \(S\) is an ordered sequenceof all elements of \(S\) such that each element appears exactly once
    • example: The set \(\{a,b,c\}\) has six possible permutations:
      • \(abc, acb, bac, bca, cab, cba\)
  • The number of permutations of a set with \(n\) elements is \(n!\)
    • for the first element in the sequence, you have \(n\) choices
    • for the second element, you have \((n-1)\) choices, and so on
    • hence number of choices = \(n*(n-1)*\cdots*1 = n!\)

Permutation

3.1 - Permutations and Combinations

Permutation

  • Given a set of size \(n\), if you only want permutations of length \(k\) (called \(\text{$k$-permutation}\)) then:
    • for the first element in the sequence, you have \(n\) choices
    • for the second element, you have \((n-1)\) choices, 
    • but you only want \(k\) elements, so you stop after choosing \(k\) elements


      (for the last element, you have \(n-k+1\) choices)

\[n * (n-1) * \cdots * (n-k+1) = \frac{n!}{(n-k)!}\]

3.1 - Permutations and Combinations

Permutation

  • Example: Given the set\(\{a,b,c,d\}\), how many permutations of 3 elements? 
    • 4 choices for the first element, 3 choices for the second one, and 2 choices for the last, i.e. a total of 24 permutations:
      • \(abc,acb,bac,bca,cab,cba\)
      • \(abd,adb,bad,bda,dab,dba\)
      • \(acd,adc,cad,cda,dac,dca\)
      • \(bcd,bdc,cbd,cdb,dbc,dcb\)

3.1 - Permutations and Combinations

Combination

  • Given a set \(S\), how many combinations of unique elements in the set can you make?
    • in permutation, the order matter: \(abc\), \(acb\), \(bca\), \(bac\), \(cab\), \(cba\) are all different
    • but if we just want a combination of 3 elements where ordering does not matter, then there is only one: \(abc\) 
  • For each combination of \(k\) elements, there are \(k!\) permutations
    • so if you want the number of combinations of \(k\) elements, divide the number of permutations with \(k!\)

3.1 - Permutations and Combinations

Combination

  • \(abc,acb,bac,bca,cab,cba\)
  • \(abd,adb,bad,bda,dab,dba\)
  • \(acd,adc,cad,cda,dac,dca\)
  • \(bcd,bdc,cbd,cdb,dbc,dcb\)
  • \(\{a,b,c\}\)
  • \(\{a,b,d\}\)
  • \(\{a,c,d\}\)
  • \(\{b,c,d\}\)
  • Thus, the number of combination of \(k\) elements is 

\[\frac{n!}{k!(n-k)!} = \binom{n}{k}\]

3.1 - Permutations and Combinations

Combination

1                4                6                 4               1

a b c d

a b c d

a b c

a b d

a c d

b c d

a b

a c

a d

b c

b d

c d

a

b

c

d

3.1 - Permutations and Combinations

Combination

1                4                6                 4               1

1                3                 3               1               

\(2^3 = 8\)

1                5               10               10              5               1

\(2^5 =32\)

\(2^4 = 16\)

a b c d

\(\frac{0}{1}\)

  • The total number of combinations is always \(2^n\) because for each item, you can either choose it, or not choose it

a d 

b c d 

  c 

1 0 0 1

\(\frac{0}{1}\)

\(\frac{0}{1}\)

\(\frac{0}{1}\)

0 1 1 1

0 0 1 0

3.1 - Permutations and Combinations

Combination

\[(x+y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^3\]

\[(x+y)^n = \sum_{k=0}^n \binom{n}{k} x^ky^{n-k}\]

  • You also use this for binomial expansion

\[(x+y)^3 = x^3 + 3x^2y + 3x2y^2 + 1y^3 \]

\[(x+y)^2 = x^2 + 2xy + y^2 \]

3.1 - Permutations and Combinations

Example

  • Question: In how many ways can you choose three distinct numbers from the set \(\{1,2,\dots,99\}\) so that their sum is even?​
  • Case 1: All numbers are even
    • \(\binom{49}{3}\) possibilities since there are 49 even numbers
  • Case 2: Two odd numbers, one even number
    • \(\binom{50}{2} * 49\)
  • Total answer: \(\binom{49}{3} + 49 * \binom{50}{2}\)

3.1 - Permutations and Combinations

Example

  • Question: How many ways can you seat \(n\) guests on a circular dining table? (assume two seatings arrangements are the same if one can be rotated from the other)

\((n-1)!\) ways            (why not \(n!\) ?)

A

B

C

D

E

A

B

C

D

E

3.1 - Permutations and Combinations

Final remarks

  • The most important takeaway from permutations and combinations is that you understand that the total number of permutation is \(n!\) and the total number of combinations is \(2^n\)
    • ​this is where we get \(O(n!)\) and \(O(2^n)\)

3.1 - Permutations and Combinations