Find the total number of permutations of the letters of each of the words given below:
(i) APPLE
(ii) ARRANGE
(iii) COMMERCE
(iv) INSTITUTE
(v) ENGINEERING
(vi) INTERMEDIATE
To find: number of permutations of the letters of each word
Number of permutations of n distinct letters is n!
Number of permutations of n letters where r letters are of one kind, s letters of another
kind, $t$ letters of a third kind and so on $=\frac{n !}{r ! s ! t ! \ldots}$
(i) Here $n=5$
P is repeated twice
So the number of permutations $=\frac{5 !}{2 !}=5 \times 4 \times 3=60$
(ii) Here $n=7$
$A$ is repeated twice, and $R$ is repeated twice
So, the number of permutations $=\frac{7 !}{2 ! .2 !}=\frac{7 \times 6 \times 5 \times 4 \times 3}{2}=1260$
(iii) Here $n=8$
$M$ and $E$ are repeated twice
So, the number of permutations $=\frac{8 !}{2 ! 2 !}=\frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2}{4}=10080$
(iv) Here $n=9$
$\mathrm{I}$ is repeated twice, $\mathrm{T}$ is repeated thrice
So, the number of permutations $=\frac{9 !}{2 ! 3 !}=30240$
(v) Here $n=11$
$\mathrm{E}, \mathrm{N}$ is repeated thrice, $\mathrm{I}, \mathrm{G}$ are repeated twice
So the number of permutations $=\frac{11 !}{3 ! 3 ! 2 ! 2 !}=277200$
(vi) Here $n=12$
$I$ and $T$ are repeated twice, $E$ is repeated thrice
So, the number of permutations $=\frac{12 !}{2 ! 2 ! 3 !}=19958400$