(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