Five letters F, K, R, R and V one in each were purchased from a plastic warehouse. How many ordered pairs of letters, to be used as initials, can be formed from them?
(i) the number of initials is 1
In this case, all letters have one chance (i.e. letters F, K, R, V ).
Formula:
Number of permutations of n distinct objects among r different places, where repetition is not allowed, is
P(n,r) = n!/(n-r)!
Therefore, a permutation of 4 different objects in 1 place is
P(4,1) = 
= 
= 
= 4.
So no of ways is 4 .
(ii) the number of initials is 2
There are two cases here
(a) When two R do not occur in initials
Formula:
Number of permutations of n distinct objects among r different places, where repetition is not allowed, is
P(n,r) = n!/(n-r)!
Therefore, a permutation of 4 different objects in 2 places is
P(4,2) = 
= 
= 
= 12.
A number of arrangements here are 12.
(b) When two R occurs in initials
When two R are chosen then 1 pair is included twice.
Selection of 0 letters remaining from 3 letters can be done in P(3,0) ways.
Formula:
A number of permutations of n objects in which p objects are alike of one kind are =n!/p!
Selections = P(3,0) × 
= 
× 
= 1
Therefore, the total number of pairs 13.
(iii) the number of initial is 3
(a) two R do not occur in initials
Formula:
Number of permutations of n distinct objects among r different places, where repetition is not allowed, is
P(n,r) = n!/(n-r)!
Therefore, a permutation of 4 different objects in 3 places is
P(4,3) = 
= 
= 
= 24.
A number of arrangements here are 24.
(b) two R occurs in initials
When two R are chosen then 1 pair is included twice.
Selection of 1 letter from the remaining 3 letters is P(3,1)
Formula:
A number of permutations of n objects in which p objects are alike of one kind = n!/p!
Selections = P(3,1) × 
= 
× 
= 9
total number of arrangements for 3 initials are 33
(iv) The number of initials is 4
(a) Two R do not occur in initials
Formula:
Number of permutations of n distinct objects among r different places, where repetition is not allowed, is
P(n,r) = n!/(n-r)!
Therefore, a permutation of 4 different objects in 4 places is
P(4,4) = 
= 
= 
= 24.
A number of arrangements here are 24.
(b) Two R occurs in the initials
When two R are chosen then 1 pair is included twice.
Selection of 2 letters from the remaining 3 letters is P(3,2)
Formula:
A number of permutations of n objects in which p objects are alike of one kind = n!/p!
Selections = P(3,2) × 
= 
× 
= 36
total number of arrangements for 4 initials are 60
(v) The number of initials is 5
Formula:
A number of permutations of n objects in which p objects are alike of one kind = n!/p!
Selections = 
= 60.
Total number of arrangements are 4 + 13 + 33 + 60 + 60 = 170