Choosing 3 roles from 10 contestants The question is:
For an association, a president, vice president and secretary, all different, are to be chosen from 10 contestants. How many different choices are possible if Ramu and Ravi will not associate together?
The answer provided is:
Without Ramu and Ravi, the number of choices is $8 \times 7 \times 6$ and with one of them in association the number of choices is $3 \times 2 \times 8 \times 7$. So the total number of choices:
$(8 \times 7 \times 6) + (3 \times 2 \times 8 \times 7) = 672$.
My answer is:
Case 1, Both Ramu and Ravi are not chosen:
8P3 = 336
Case 2, Either Ramu or Ravi is not chosen:
$(8 \times 7 \times 3!) \times 2 = 672$
Total different choices: $672 + 336 = 1008$
Why shouldn't I multiply by 2? My thought process is... $(8 \times 7 \times 3!)$ only takes into account that if one of them, say Ramu, was chosen. But Ravi can also be chosen and I have to take that case into account too?
If someone could explain to me why I shouldn't multiply by 2... that would be great, thanks!
 A: It's not the final $\times 2$ that is wrong, it is the $8\times 7\times 3!$, which should be $8\times 7\times 3$. Once you choose which position is Ramu/Ravi ($3$ options), there are ${}^8P_2=8\times 7$ to choose people for the other two positions in order. 
A: There are $8\times 7$ solutions with Ramu president. Same for vice-president, same for secretary. Repeat with Ravi and you get the $6 \times 8\times 7$.
A: Exactly one of Ramu or Ravi is selected:  We must select either Ramu or Ravi, select two of the other eight people, then assign positions to the three selected people.
$$\binom{2}{1}\binom{8}{2}3!$$

Why shouldn't I multiply by $2$? My thought process is... $(8 \times 7 \times 3!)$ only takes into account that if one of them, say Ramu, was chosen.  But Ravi can also be chosen and I have to take case into account too?

The factor of $2$ in your calculation refers to the choice between Ramu and Ravi.  The factor of $3!$ is the number of ways we can assign the three people we have selected to the available positions.  In multiplying $8$ by $7$, you are counting all ordered selections of two of the other eight people.  However, if the positions have not yet been assigned, choosing Madhuri and Juhi to serve with Ramu has the same effect as choosing Juhi and Madhuri to serve with Ramu.  Hence, the number of distinguishable ways of choosing two of the other eight people to serve with Ramu (or Ravi) is 
$$\binom{8}{2} = \frac{8 \cdot 7}{2}$$ 
