Question

Consider the word ${}^{\prime }TELANGAN{A}^{\prime }$, then the number of possible words

• A. without any restrictions is $\frac{9!}{2!3!}$
• B. If word starts with consonant is $4\times \frac{8!}{2!3!}$
• C. when the word starts and end with any vowel is $7!$
• D. If word starts with $T$ and ends with $G$ is $420$

Answer 1

Answer: (A), (C), (D)

The correct options are
A without any restrictions is $\frac{9!}{2!3!}$
C when the word starts and end with any vowel is $7!$
D If word starts with $T$ and ends with $G$ is $420$

There is total $9$ letters, there is $3A^{\prime }s$ and $2N^{\prime }s$
$\therefore$Total number of possible permutations is $\frac{9!}{2!3!}$

If word starts with $T,L,G$, then the number of possible permutations is $\frac{8!}{2!3!}$

If word starts with $N$, then the number of possible permutations is $\frac{8!}{3!}$

Total ways are $3\times \frac{8!}{2!3!}+\frac{8!}{3!}$

When the word starts and end with any vowel
If the end letters are $2A^{\prime }s$
Total ways are $=\frac{7!}{2!}$
If the end letters are $E$ and $A$
Total ways are $=2\times \frac{7!}{2!2!}$
$\therefore$ ​​​​​​​​​​​​​​Total ways are $=\frac{7!}{2!}+2\times \frac{7!}{2!2!}=7!$

When word starts with $T$ and ends with $G$, then the letters in between can be arranged as $\frac{7!}{2!3!}=420$