Permutations and Combinations

1
Permutations and Combinations

• Rosen 4.3

2
Permutations

• A permutation of a set of distinct objects is an
  ordered arrangement these objects.
• An ordered arrangement of r elements of a set is
  called an r-permutation.
• The number of r-permutations of a set with n
  elements is denoted by P(n,r).
• A 1,2,3,4 2-permutations of A include 1,2
  2,1 1,3 2,3 etc

3
Counting Permutations

• Using the product rule we can find P(n,r)
• n(n-1)(n-2) (n-r1)
• n!/(n-r)!
• How many 2-permutations are there for the set
  1,2,3,4? P(4,2)

4
Combinations

• An r-combination of elements of a set is an
  unordered selection of r element from the set.
  (i.e., an r-combination is simply a subset of the
  set with r elements).
• Let A1,2,3,4 3-combinations of A are
• 1,2,3, 1,2,4, 1,3,4, 2,3,4(same as
  3,2,4)
• The number of r-combinations of a set with n
  distinct elements is denoted by C(n,r).

5
Example

• Let A 1,2,3
• 2-permutations of A are 1,2 2,1 1,3 3,1
  2,3 3,2
• 6 total. Order is important
• 2-combinations of A are 1,2, 1,3, 2,3
• 3 total. Order is not important
• If we counted the number of permutations of each
  2-combination we could figure out P(3,2)!

6
How to compute C(n,r)

• To find P(n,r), we could first find C(n,r), then
  order each subset of r elements to count the
  number of different orderings. P(n,r)
  C(n,r)P(r,r).
• So C(n,r) P(n,r) / P(r,r)

7
A club has 25 members.

• How many ways are there to choose four members of
  the club to serve on an executive committee?
• Order not important
• C(25,4) 25!/21!4! 25242322/4321
  252322 12,650
• How many ways are there to choose a president,
  vice president, secretary, and treasurer of the
  club?
• Order is important
• P(25,4) 25!/21! 303,600

8
The English alphabet contains 21 consonants and 5
vowels. How many strings of six lower case
letters of the English alphabet contain

• exactly one vowel?
• exactly 2 vowels
• at least 1 vowel
• at least 2 vowels

9
The English alphabet contains 21 consonants and 5
vowels. How many strings of six lower case
letters of the English alphabet contain

• exactly one vowel?
• Note that strings can have repeated letters!
• We need to choose the position for the vowel
• C(6,1) 6!/1!5! This can be done 6 ways.
• Choose which vowel to use.
• This can be done in 5 ways.
• Each of the other 5 positions can contain any of
  the 21 consonants (not distinct).
• There are 215 ways to fill the rest of the
  string.
• 65215

10
The English alphabet contains 21 consonants and 5
vowels. How many strings of six lower case
letters of the English alphabet contain

• exactly 2 vowels?
• Choose position for the vowels.
• C(6,2) 6!/2!4! 15
• Choose the two vowels.
• 5 choices for each of 2 positions 52
• Each of the other 4 positions can contain any of
  21 consonants.
• 214
• 1552214

11
The English alphabet contains 21 consonants and 5
vowels. How many strings of six lower case
letters of the English alphabet contain

• at least 1 vowel
• Count the number of strings with no vowels and
  subtract this from the total number of strings.
• 266 - 216

12
The English alphabet contains 21 consonants and 5
vowels. How many strings of six lower case
letters of the English alphabet contain

• at least 2 vowels
• Compute total number of strings and subtract
  number of strings with no vowels and the number
  of strings with exactly 1 vowel.
• 266 - 216 - 65215

13
Corollary 1 Let n and r be nonnegative integers
with r ? n. Then C(n,r) C(n,n-r)

• Proof
• C(n,r) n!/r!(n-r)!
• C(n,n-r) n!/(n-r)!(n-(n-r))! n!/r!(n-r)!

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Binomial Coefficient

• Another notation for C(n,r) is . This
  number is also called a binomial coefficient.
• These numbers occur as coefficients in the
  expansions of powers of binomial expressions such
  as (ab)n.

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Pascals Identity

• Let n and k be positive integers with n ? k.
  Then C(n1,k) C(n, k-1) C(n,k).
• Proof

16
Let n be a positive integer. Then

• Proof We know from set theory that the number of
  subsets in a set of size n is 2n. We also know
  that C(n,k) is the number of subsets of a set of
  size n that are of size k.
• counts the number of
  subsets
• of every size from 0 (empty set) to n.
  Therefore the sum must add up to 2n.

17
Vandermondes Identity
Proof Suppose there are n items in one set and m
items in a second set. Then the total number of
ways to pick r elements from the union of these
sets is C(mn,r). Another way to pick r elements
from the union is to pick k elements from the
first set and then r-k elements from the second
set, where 0 ? k ? r. There are C(n,k) ways to
pick the k elements from the first set and
C(m,r-k) ways to pick the rest of the elements
from the second set.
18

• Proof Suppose there are n items in one set and m
  items in a second set. Then the total number of
  ways to pick r elements from the union of these
  sets is C(mn,r).
• Another way to pick r elements from the union is
  to pick k elements from the first set and then
  r-k elements from the second set, where 0 ? k ?
  r. For any k,there are C(n,k) ways to pick the k
  elements from the first set and C(m,r-k) ways to
  pick the rest of the elements from the second
  set. By the product rule there are
  C(m,r-k)C(n,k) ways to pick r elements for a
  particular k. For all possible values of k

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Pascals Triangle
1
1
1
1
2
1
1
3
3
1
1
4
6
4
1
nth row, Cnk k 0, 1, , n
20
Binomial Theorem

• Let x and y be variables and let n be a positive
  integer. Then