All the letters of the word $\text{NUMBER}$ are arranged in different possible ways. What are the number of arrangements possible so that no two words are repeated?

The correct option is A $6!$
There are $6$ letters in the word $\text{NUMBER}$ which are arranged in different possible ways so that none of the word is repeated.

We have to find number of all those possible arrangements, which is similar to filling of $6$ vacant places without repetition or arrangements of $6$ different things taken all at a time.


So, number of possible ways will be $={}^6{P}_6=\frac{6!}{(6-6)!}=\frac{6!}{0!}=6!.$