It should be 7
See here is 3 Os
So, here the trick lies
First from 3 Os, u r selecting 2 O's =$\binom{3}{2}\times \frac{2!}{3!}$=1..............................i
[Now why this 3!
Here is the reason
when there is 2 same letter we divide it by 2! ,
why this 2!, because say for $O_{1}O_{2}$
it can be arranged like $O_{1}O_{2} and O_{2}O_{1}$
That is why we divide by 2!
Now in case of 3!
there is$O_{1}O_{2}O_{3}$
Now we can select 2 among these 3 like
$O_{1}O_{2}$,$O_{1}O_{3}$,$O_{2}O_{1}$ ,$O_{2}O_{3}$,$O_{3}O_{1}$,$O_{3}O_{2}$
So 6 arrangement is possible
That is why we divide by 3!]
I guess upto this u r clear
Now it is easy
Select 1 O's among 3 O's and from N and R select 1 letter = $\binom{3}{1}\binom{2}{1}\frac{2!}{3}$=4 [divide 3 because we can select any one O's from 3 O's]...............................ii
and lastly select only N and R and arrange them=$\binom{2}{2}\times 2!$=2........................iii
Now adding i , ii and iii we get 7