Unordered Selection - Discrete Mathematics - Lecture Slides, Slides of Discrete Mathematics

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CSE115/ENGR160 Discrete Mathematics

6.3 Permutations & Combinations

• Counting:

• Find out the number of ways to select a particular

number of elements from a set

• Sometimes the order of these elements matter

• Example:

• How many ways we can select 3 students from a

group of 5 students?

• How many different ways they stand in line for

picture?

Permutation

• How many ways can we arrange all 5 in a line

for a picture?

• By product rule, we have 5x4x3x2x1=120 ways

to arrange all 5 students in a line for a picture

Permutation

• A permutation of a set of distinct objects is an

ordered arrangement of these objects

• An ordered arrangement of r elements of a set is

called an r-permutation

• The number of r-permutation of a set with n element

is denoted by P(n,r). We can find P(n,r) using the

product rule

• Example: Let S={1, 2, 3}. The ordered arrangement 3,

1, 2 is a permutation of S. The ordered arrangement

3, 2, is a 2-permutation of S

r-permutation

• Theorem 1: If n is a positive and r is an integer with

1 ≤r≤n, then there are

P(n,r)=n(n-1)(n-2)…(n-r+1)

r-permutations of a set with n elements

• Proof: Use the product rule, the first element can

be chosen in n ways. There are n-1 ways to chose

the 2 nd^ element. Likewise, there are n-2 ways to

choose 3 rd^ element, and so on until there are

exactly n-(r-1)=n-r+1 ways to choose the r-th

element. Thus, there are n∙(n-1)∙(n-2)… ∙(n-r+1)

r-permutations of the set 7

r-permutation

• Note that p(n,0)=1 whenever n is a nonnegative

integer as there is exactly one way to order zero

element

• Corollary 1: If n and r are integers with 0≤r≤n, then

P(n,r)=n!/(n-r)!

• Proof: When n and r are integers with 1 ≤r≤n, by

Theorem 1 we have

P(n,r)=n(n-1)…(n-r+1)=n!/(n-r)!

• As n!/(n-0)!=1 when n is a nonnegative integer,

we have P(n,r)=n!/(n-r)! also holds when r=

Example

• How many permutations of the letters

ABCDEFGH contain string ABC?

• As ABC must occur as a block, we can find the

answer by finding the permutations of 6

letters, the block ABC and the individual

letters, D,E,F,G, and H. As these 6 objects must

occur in any order, there are 6!=

permutations of the letters ABCDEFGH in

which ABC occurs in a block

Combinations

• How many different committees of 3 students can be formed

from a group of 4 students?

• We need to find the number of subsets with 3 elements from

the set containing 4 students

• We see that there are 4 such subsets, one for each of the 4

students as choosing 4 students is the same as choosing one

of the 4 students to leave out of the group

• This means there are 4 ways to choose 3 students for the

committee, where th order in which these students are

chosen does not matter

Example

• Let S be the set {1, 2, 3, 4}. Then {1, 3, 4} is a

3-combination from S

• We see that C(4,2)=6, as the 2-combination of

{a, b, c, d} are 6 subsets {a, b}, {a, c}, {a, d}, {b,

c}, {b, d}, and {c, d}

r-combination

• We can determine the number of r-

combinations of a set with n elements using

the formula for the number of r-permutations

of a set

• Note that the r-permutations of a set can be

obtained by first forming r-combinations and

then ordering the elements in these

combinations

r-combination

• When computing r-combination

thus canceling out all the terms in the larger

factorial

16

r

n n n r r

P n r r n r

C n r n = = − − + −

Example

• How many poker hands of 5 cards can be dealt

from a standard deck of 52 cards? Also, how

many ways are there to select 47 cards from a

standard deck of 52 cards?

• Choose 5 out of 52 cards: C(52,5)=52!/(5!47!)=

(52x51x50x49x48)/(5x4x3x2x1)=26x17x10x

x12=2,598,

• C(52,47)=52!/(47!5!)=2,5,98,

Combinatorial proof

• A combinatorial proof of an identity is a proof

that uses counting arguments to prove that

both sides of the identity count the same

objects but in different ways

• Proof of Corollary 2: Suppose that S is a set

with n elements. Every subset A of S with r

elements corresponds to a subset of S with n-r

elements, i.e.,. Thus, C(n,r)=C(n,n-r)

19

A

Example

• How many ways are there to select 5 players

from a 10-member tennis team?

• Choose 5 out of 10 elements, i.e., C(10,

• How many bit strings of length n contain

exactly r 1s?

• This is equivalent to choose r elements from n

elements, i.e., C(n,r)