Question

If all permutations of the letters of the word AGAIN are arranged in the order as in a dictionary. What is the ${49}^{th}$ word?

Answer 1

Solution :-

Given, the letter of the word AGAIN.

Now we will arranged in alphabetically order. AGAIN

starting with 'A' and arranging the other four letters

we have = 4: ways = 24 ways

starting with 'G', arranging the other four letter = $\frac{4!}{2!}=12$ ways

starting with 'I' we have = $\frac{4!}{2!}=12$ = 12 ways.

so far we have $24+12+12=48$ words

$\therefore$ the ${49}^{th}$ word will start from N and is NAAGI