How many $4$ letter words can be formed from the word "CORONAVIRUS". Letters: $C$, $O$ ($2$ times), $R$ ($2$ times), $N$, $A$, $V$, $I$, $U$ and $S$.
The number of $4$ letter words from $C$, $N$, $A$, $V$, $I$, $U$ and $S$ is $7\times 6\times 5\times 4=840$.
The number of $4$ letter words from two $O$ or two $R$ while other $2$ letters are different is $\displaystyle \binom{2}{1}\times \binom{8}{2}\times \frac{4!}{2!}=672$.
The number of $4$ letter words from two $O$ and two $R$ together is $\displaystyle \frac{4!}{2!\times 2!}=6$
Total number of ways should be $840+672+6=1518$.
But this is not the right answer given. What am I doing wrong? What cases I am missing here? Please help!!!
Thanks in advance!!!

We can consider the following cases:
Case 1. No letter is repeated.
In this case, we obtain $P_4^9=3024$ different words.
Case 2. A word contains 2 R's or 2 O's
Case 2.1 A word contains 2 R's and 0 O's
In this case, we obtain $\binom{7}{2} \times \frac{4!}{2!}=252$ different words.
Case 2.2 A word contains 2 R's and 1 O's
In this case, we obtain  $\binom{7}{1} \times \frac{4!}{2!}=84$ different words.
Case 2.3 A word contains 2 R's and 2 O's
In this case, we obtain  $\frac{4!}{2!2!}=6$ different words.
Case 2.4 A word contains 0 R's and 2 O's
In this case, we obtain $\binom{7}{2} \times \frac{4!}{2!}=252$ different words.
Case 2.5 A word contains 1 R's and 2 O's
In this case, we obtain  $\binom{7}{1} \times \frac{4!}{2!}=84$ different words.
The total number of 4-letter words is $15120+252+84+6+252+84=3702$.
