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Permutations
☼ PERMUTATIONS : In the above example, 2 objects were selected from 3 objects in such a way that the order of the arrangements is important.
Now, if there were n = 10 objects and r = 3 objects were to be selected, then we could make the selection in 10 9
8 ways.
This can also be written as =
=
=
Thus, we can say that the permutation of n objects taken r at a time is given by,
■ =
= n
( n – 1 )
( n – 2 )
…
{ n – ( r – 1 ) } ; 0
r
n
■ = 1,
= n,
= n !
■ = 8
7
6 = 336,
= 12
11
10
9
8 = 95040,
= 5,
= 1
e.g. (i) How many seven-letter words can be formed from the letters of the word ‘HEXAGON’ ? How many of them begin with E and end with N ?
Soln. The 7 places of letters of the word can be filled by 7 different letters of given word by = 7 ! = 5040 ways.
& The no. of words begin with E and end with N = = 5 ! = 120
(2) How many six-letter words can be formed from the letters of the word ‘LADDER’ ?
Soln. Total 6 letters are in the given word. And ‘D’ is repeated twice.
No. of required no. of words = =
= 360.
(2) How many six-letter words can be formed from the letters of the word ‘NAYANA’ ?
Soln. Total 6 letters are in the given word. And ‘N’ repeated twice whereas ‘A’, repeated thrice.
No. of required no. of words =
=
= 60.
