The document discusses permutations and combinations. It defines a permutation as an ordered arrangement of objects from a set, while a combination is an unordered selection of objects from a set. It provides formulas for calculating the number of permutations P(n,r) and combinations C(n,r) of choosing r objects from a set of n. Examples are given to demonstrate calculating permutations and combinations in different scenarios.

2. Permutations and
Combinations
Name: Ghufran Hasan
CMS: 11432
Aqib javaid :13765
Abdul waheed : 14200

3. Permutations
 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-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…

4. Counting Permutations
 Using the product rule we can find P(n,r)
= n*(n-1)*(n-2)* …*(n-r+1)
= n!/(n-r)!
How many 2-permutations are there for the set {1,2,3,4}?
P(4,2)
12
!2
!4
1*2
1*2*3*4
3*4 ===

5. How many Permutations?
 Consider four objects {A,B,C,D}
 There are 4 choices for the first slot.
 There are 3 choices for the second slot.
 There are 2 choices for the third slot.
 There is 1 choice for the last slot.

6. 4 x 3 x 2 x 1 = 24
Permutations
ABCD ABDC ACBD
ACDB ADBC ADCB
BACD BADC BCAD
BCDA BDAC BDCA
CABD CADB CBAD
CBDA CDAB CDBA
DABC DACB DBAC
DBCA DCAB DCBA

7. Generalization
There are 4! ways to
arrange 4 items.
There are n! ways to
arrange n items.

8. Permutation Formula
In how many ways may r items be selected out of a set
of n items where order matters
)!(
!
),(
rn
n
rnPPrn
−
==

9. Permutation Example
Selecting 3 items out of a set of 5
We have 5 choices for the first item.
We have 4 choices for the second item.
We have 2 choices for the third item.
5 x 4 x 3 = 60 Permutations

10. Calculations
)!35(
!5
)3,5(
!2
!5
12
12345
345
35
−
==
=
⋅
⋅⋅⋅⋅
=⋅⋅
CC

11. Permutation Formula
)!(
!
),(
rn
n
rnPPrn
−
==

12. 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 A={1,2,3,4} 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).

13. Evaluating
In how many ways may 3 items be selected from a set
of 5 without regard to order?
rnC

14. We already know that there 60
permutations of these items.
For each set of three, there are
3! or 6 arrangements.
A B C A C B B A C
B C A C A B C B A
All of these are really the same.

15. Our actual answer is 10.
Consider the set {A,B,C,D,E}
These are the combinations.
A,B,C A,B,D A,B,E A,C,D A,C,E A,D,E B,C,D
B,C,E B,D,E C,D,E
10
6
60
!3
35
==
C

16. Combination Formula
)!(!
!
),(
rnr
n
rnCCrn
−
==

17. 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)
)!(!
!
!)!(
)!(!
)!(
!
)!(
!
rnr
n
rrn
rrn
rr
r
rn
n
−
=
−
−
=
−
−
=

18. 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)!

19. 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! = 25*24*23*22/4*3*2*1
=25*23*22 = 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

20. Thank You