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 50th word?

Total number of words:
Number of ways to arrange 5 elements = 5!.
When a permutation includes IDENTICAL elements, we must divide by the number of ways to ARRANGE the identical elements.
The reason is that the permutation doesn't change when the identical elements swap positions.
Number of ways to arrange the two identical A's = 2!.
Thus:
Total possible permutations = 5!/2! = 60.

50th word:
Work BACKWARDS from the END of the list.
When the 60 words are listed in alphabetical order, those at the end of the list will all begin with N.
If the first letter is N, the number of ways to arrange the remaining 4 letters AAGI = 4!/2! = 12.
Thus, there are 12 words that begin N.
These 12 words constitute the LAST 12 WORDS in the list:
Words 49 through 60.

Thus:
Word 49 is the FIRST word that starts with N:
NAAGI.
Word 50 is the SECOND word that starts with N:
NAAIG.

HI

Check my previous post

https://www.beatthegmat.com/p-c-t202402.html

U can get it by forward counting

words start with A = 24 (1x4x3x2x1)
words start with G = 12 (1x4x3x2x1/2!)
words start with I = 12 (1x4x3x2x1/2!)
so the 49th word in dictonery = NAAGI and 50th will be NAAIG