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Permutations and Combinations

Learning Objectives

- What are permutations.
- What are combinations.
- How to calculate binomial coefficients.
- What is the binomial theorem.
- Counting examples.

Permutations and Combinations

- Urn models
- We are given set of n objects in an urn (dont

ask why its called an urn - probably due to

some statistician years ago) . - We are going to pick (select) r objects from the

urn in sequence. After we choose an object - we can replace it-(selection with replacement)
- or not -(selection without replacement).
- If we choose r objects, how many different

possible sequences of r objects are there? - Does the order of the objects matter or not?

Permutations and Combinations

- PermutationsSelection without replacement of r

objects from the urn with n objects. - A permutation is an arrangement.
- Order matters .
- After selecting the objects, two different

orderings or arrangements constitute different

permutations.

Permutations and Combinations

- Choose the first object n ways,
- Choose the second object (since selection is

without replacement) (n - 1) ways, - ...
- Choose the rth object (n - r 1) ways.
- By the rule of product, The number of

permutations of n things taken r at a time - P(n,r) n(n - 1)(n - 2) . . . (n - r 1)
- Note

Permutations and Combinations

- ExampleLet A and B be finite sets and let A

? B .Count the number of injections from A

to B.Note there are no injections if A gt B

(why?) - There are P( B , A ) injections
- We order the elements of A, a1, a2, . . . and

assume the urn contains the set B. - There are B ways to choose the image of a1,

B - 1 ways to choose the image of a2, and so

forth. - Selection is without replacement. Otherwise we do

not construct an injection.

Permutations and Combinations

- Combinations
- Selection is without replacement but order does

not matter . - It is equivalent to selecting subsets of size r

from a set of size n. - Divide out the number of arrangements or

permutations of r objects from the set of

permutations of n objects taken r at a time - The number of combinations of n things taken r at

a time

Permutations and Combinations

- Other names for C(n, r)
- n choose r
- The binomial coefficient
- Example
- How many subsets of size r can be constructed

from a set of n objects? - The answer is clearly C(n, r) since once we

select the objects (without replacement) the

order doesn't matter.

Permutations and Combinations

- CorollaryProof
- If we count the number of subsets of a set of

size n, we get the cardinality of the power set.

Permutations and Combinations

- ExampleSuppose you flip a fair coin n times.

How many different ways can you get - no heads? C(n, 0)
- exactly one head? C(n, 1)
- exactly two heads? C(n, 2)
- exactly r heads? C(n, r)
- at least 2 heads? 2 n - C(n, 0) - C(n, 1)

Permutations and Combinations

- Pascal's IdentityProof
- We construct subsets of size k from a set with n

1 elements given the subsets of size k and k-1

from a set with n elements. - The total will include
- all of the subsets from the set of size n which

do not contain the new element C(n, k), - plus
- the subsets of size k - 1 with the new element

added C(n, k-1).

Permutations and Combinations

- It produces

Permutations and Combinations

- A good way to evaluate C(n, r) for large n and r

(to avoid overflow). - ExampleHow many bit strings of length 4 have

exactly 2 ones (or exactly 2 zeros)?

Permutations and Combinations

- AnalysisWe solve the problem by determining the

positions of the two ones in the bit string. - place the first one - 4 possibilities
- place the second one - 3 possibilities
- Hence it appears that we have (4)(3) 12

possibilities. - We enumerate them to make sure
- 0011, 0101, 1001, 0110, 1010, 1100.
- There are actually only 6 possibilities. What is

wrong?

Permutations and Combinations

- The answer would be correct if we had two

different objects to place in the string. - For example, if we were going to place an a and

a b in the string we would have00ab, 00ba,

0a0b, 0b0a, a00b, b00a,and so forth for a total

of 12. - But......the objects (1 and 1) are the same so

the order is not important ! - Divide through by the number of orderings 2!

2. Therefore the answer is 12/2 6.

Permutations and Combinations

- ExampleHow many bit strings of length 4 have at

least 2 ones?AnalysisTotal the number of

strings that have - zero 1s 1
- one 1 4Total 24 - 5 11.
- If the universe is the bit strings of length 4,

what is the complement of the above set? - What is its cardinality?

Counting Examples

- An office building contains 27 floors and 37

offices on each floor. How many offices are in

the building ? - Answer 27 37 999 (product rule)

Counting Examples

- How many bit strings are there of length 6 or

less ? - Answer sum rule
- length 1 2
- length 2 4
- length 3 8
- length 4 16
- length 5 32
- length 6 64
- total 2 4 8163264 126

Counting Examples

- How many strings of four decimal digits
- do not contain the same digit twice
- 10987 5040
- end with an even digit
- 1010105 5000
- have exactly three digits that are 9s
- 94 36

Counting Examples

- How many different functions are there from a set

with 10 elements to a set with - one element 1
- two elements 2 10
- three elements 3 10

Counting Examples

- How many bit strings of length seven begin with

three 0s or end with three 1s? - Principle of inclusion-exclusion
- begin with three 0s 2 4 16
- end with three 1s 2 4 16
- intersection 2
- total 16 16 - 2 30

Pigeon-Hole Principle Examples

- Show that among any group of five (not

necessarily consecutive integers), there are two

with the same remainder when divided by 4. - Pigeon-hole principle with 4 boxes (there are 4

different remainders of division by 4) - 41 5
- in any 5 integers, there are at least two with

the same remainder when divided by 4

Pigeon-Hole Principle Examples

- Suppose that there are 9 students in a class.
- Show that the class must have at least five male

students or at least five female students. - Generalized pigeon-hole principle 2 boxes,

?9/2? 5 - Show that the class must have at least three male

students or at least seven female students. - Either nb males gt2.
- Or nb males lt 2, and then nb females gt 7.

Permutation and Combination

- List all permutations of a, b, c
- Permutations of a, b, c are bijections from and

onto this same set. There are 3! 6

permutations. - a, b, c a, c, b b, a, c b, c, a

c, a, b c, b, a

Permutation and Combination

- A group contains n men and n women. How many ways

are there to arrange these people in a row if the

men and women alternate ?M W M W for M

n ! P(n, n) for W n ! P(n,

n) sub-total n! 2WMWM same n! 2total 2

n! 2

Permutation and Combination

- In how many ways can a set of five letters be

selected from the English alphabet. - We suppose that letters cannot be repeated
- If the order matters (such as in a word) P(26,

5) 26.25.24.23.22 7893600Example abcdef,

bacdef, ... - If the order does not matter (such as in a

set) C(26,5) P(26,5) / 5! 65780 Example

a, b, c, d, e, f

Permutation and Combination

- If letters can be repeated
- If the order matters 26 5 11881276
- If the order does not matter more complex

C(26 5 - 1, 5) C(30, 5) 142506 - Number of groups of p objects (not necessarily

different) with repetition taken from a set of m

objects without taking the order into account

C(m p - 1, p)

Permutation and Combination

- A department contains 10 men and 15 women. How

many ways are there to form a committee with 6

members if it must have more women than men. - Committees are not ordered and no duplicates are

allowed. - 6 women C(15, 6) . C(10,0)
- 5 women C(15,5) . C(10,1)
- 4 women C(15,4) . C(10,2)
- total (sum rule) C(15,6) C(15,5).10

C(15,4).45

Permutation and Combination

- Give the row of Pascals triangle immediately

following 1 7 21 35 35 21 7 11 11 2

11 3 3 11 4 6 4 11 5 10 10 5

11 6 15 20 15 6 11 7 21 35 35 21 7

1 1 8 28 56 70 56 28 8 1

Permutation and Combination

- Develop (x y) 8 (x y)8 x8 8 x7 y 28

x6 y2 56 x5 y3 70 x4 y4 56 x3 y5 28 x2

y6 8 x y7 y8

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