Five letters F, K, R, R and V one in each were purchased from a plastic
Question:

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?

Solution:

(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

$P(4,1)=\frac{4 !}{(4-1) !}$

$=\frac{4 !}{3 !}=\frac{24}{6}=4 .$

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

$P(4,2)=\frac{4 !}{(4-2) !}$

$=\frac{4 !}{2 !}=\frac{24}{2}=12$

A number of arrangements here are 12 .

(b) When two R occurs in initials

When two $\mathrm{R}$ are chosen then 1 pair is included twice.

Selection of 0 letters remaining from 3 letters can be done in $\mathrm{P}(3,0)$ ways.

Formula:

A number of permutations of $n$ objects in which $p$ objects are alike of one kind are $=n ! / p !$

Selections $=P(3,0) \times \frac{2 !}{2 !}$

$=\frac{3 !}{3 !} \times \frac{2 !}{2 !}=1$

Therefore, the total number of pairs 13 .

(iii) The number of initial is 3

(a) two $\mathrm{R}$ 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

$P(4,3)=\frac{4 !}{(4-3) !}$

$=\frac{4 !}{1 !}=\frac{24}{1}=24$

A number of arrangements here are 24 .

(b) two $\mathrm{R}$ 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 $=P(3,1) \times \frac{3 !}{2 !}$

$=\frac{3 !}{2 !} \times \frac{3 !}{2 !}=9$

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

$P(4,4)=\frac{4 !}{(4-4) !}$

$=\frac{4 !}{0 !}=\frac{24}{1}=24$

A number of arrangements here are 24 .

(b) Two $\mathrm{R}$ 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 $=\mathrm{P}(3,2) \times \frac{4 !}{2 !}$

$=\frac{3 !}{1 !} \times \frac{4 !}{2 !}=36$

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 $=\frac{5 !}{2 !}=60$

Total number of arrangements are 4 + 13 + 33 + 60 + 60 = 170