Question

The number of words with or without meaning which can be made using all the letters of the word AGAIN are ...A... and if these words are written as in dictionary, then the $50^{\text{th }}$ word is ...B.... Here, A and $B$ refer to

• A. 60 , NAAIG
• B. 60, NAAGI
• C. 120 , NAAIG
• D. 120 , NAAGI

Answer 1

Answer: (A)

There are 5 letters in the word 'AGAIN', in which A appears 2 times. Therefore, the required number of words $=\frac{5!}{2!}=60$.
To get the number of words starting with $A$, we fix the letter $A$ at the extreme left position, we then rearrange the remaining 4 letters taken all at a time. There will be as many arrangements of these 4 letters taken 4 at a time as there are permutations of 4 different things taken 4 at a time. Hence, the number of words starting with $A=4!=24$. Then, starting with $G$, the number of words $=\frac{4!}{2!}=12$ as after placing $G$ at the extreme left position, we are left with the letters A, A, I and N. Similarly, there are 12 words starting with the next letter I. Total number of words so far obtained $=24+12+12=48$.
The $49^{th}$ word is NAAGI. The $50^{th}$ word is NAAIG.