MATHEMATICS
Permutations
Permutations
 Suppose now the tourist come to town where there are 4
places to visit (A, B, C, D). And because he wants to go
home soon, he can only visit 2 places. There are several
possibilities about how he visites the two places. He may
go to A first then to B, he may go to B first then to A, he
may go to A then to C, etc.
 The following diagram represents those possibilities.
So there are 12 possibilities.
We may also use the following multiplication boxes.
The box (i) represents the first place to visit and box (ii)
represents the second place.
In general, if there are n places to visit, while the tourist
can only visit r places, then the number of ways the tourist
visits the r places is:
Here, is called permutations of n to r.
And since:
prn  n(n  1)(n  2)...(n  r  1)

n(n  1)( n  2)...( n  r  1)( n  r )!
(n  r )!
n!
p 
(n  r )!
n
r
For instance, in our example above,
p rn 
4!
4! 4.3.2.1
 
 12
(4  2)! 2!
2.1
Example
1
There are 5 non collinear points. How
vectors can be formed?
Answer = 5 x 4 = 20
The following diagram shows those vectors
(A, B, C, D, E are points).
Note: notice that AB  BA, BC  CB,
Which means: the order, is important in permutations.
Example 2
 There are 7 executive, where there executive shall be chosen as
marketing manager, after sales manager, and human resources
manager. Find the number of possibilities.
 Answer:
p37  7.6.5  210
 Example 13 Prove that:
p
n 1
r
(n  1)!
(n  1)
n!


.
(n  1  r )! (n  1  r ) (n  r )!
n!
p 
(n  r )!
n
r
Subtracting
p
n 1
r
n!  n  1

p 
 r

(n  r )!  n  1  r

n
r
n! (n  1  n  1  r )

(n  r )!
(n  r  1)
n!
 r.
(n  r  1)!
 rp
n
r 1
Example 3
 If
p32 n  84n
then find
p
( n 1)
2
Answer:
2n (2n – 1) (2n – 2) = 84n
(2n – 1) (2n – 2) = 42
2n2 – 3n – 20 = 0
(2n + 5) (n – 4) = 0

n=4
p 2n1  p241  5.4  20
Thank You