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!!!
 A: In your work the first case should be 4 distinct letter words from C,N,A,V,I,U,S,O,R, (O and R included!) that is $9×8×7×6 = 3024$ ways.
Threrefore the total number of $4$-letter words is $3024+672+6=3702$.
A: The first case holds the mistake: You must have considered that to be the case where no repetitions of letters occur but have failed to account for the letters O and R  appearing just once, which you seem to have done correctly in the second case (one letter repeated).  Eg. You’ve left out words like CORN, SOUR, SCAR etc.
The correct answer would be then to choose from 9 letters to start with and then keep decreasing the multiplicand by 1:$$9\times8 \times7 \times6=3024$$ and the sum is $$3024+672+6=3702.$$
A: Four letters from the letters C,N,A,V,I,U,S,O,R can be arranged in $9\times 8\times 7\times 6=3024$ different ways.
Four letters from the letters O,O,R,R can be arranged in $\frac{4!}{2!2!}=6$ different ways.
Four letters from the letters C,N,A,V,I,U,S,R and O,O can be arranged in $\binom{8}{2} \frac{4!}{2!1!1!}=336$ different ways.
Four letters from the letters C,N,A,V,I,U,S,O and R,R can be arranged in $\binom{8}{2} \frac{4!}{2!1!1!}=336$ different ways.
Total number of different arrangment is $3702$
A: 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 $15120+252+84+6+252+84=3702$.
A: You could also use the exponential generating function:
$(1+x+x^2/2)^2(1+x)^7$
The fourth power term is $\dfrac{617x^4}{4}$.  The number of 4 letter words would be the coefficient of $x^4/4!$. So $617 \times 6=3702$.
