Find the total number of permutations of the letters of each of the words

Question:

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

Solution:

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 $=\frac{12 !}{2 ! 2 ! 3 !}=19958400$

 

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