Question

Find the number of words with or without meaning which can be made using all the letters of the word $AGAIN$. If these words are written as in a dictionary, what will be the ${50}^{th}$ word?

Answer 1

There are $5$ letters in the word AGAIN, in which $A-2$

$\therefore$ Required numbers of ways

$=\frac{5!}{2!}=\frac{5\times 4\times 3\times 2\times 1}{2\times 1}=60$

To get the number of words starting with A , fix A in first place and remaining $4$ letters can be arranged in $4!$ ways = $24$ ways

To get the number of word starting with G, fix G in first place and the remaining $4$ letters in which $A-2$

$\therefore$ Number of word starting with G $=\frac{4!}{2!}=12$

Similarly, numbers of words starting with $I=\frac{4!}{2!}=12$

Total number of words so far obtained $=24+12+12=48$

The ${49}^{th}$ word is NAAGI and ${50}^{th}$ word is NAAIG