Permutation and combination with overcounting A lady gives a dinner party to six quests. The number of ways in which they may be selected from among ten friends, if two of the friends will not attend the party together is 
(a) $112$     (b) $140$ 
(c) $164$      (d) None of these 

what i tried:
 Since two friend will not attend party together i considered them as one, so now there 9 friends.
So the number of ways $6$ people are selected from $9$ people=$\frac{9\times8\times7\times6\times5\times4}{6\times5\times4\times3\times2\times1}$=$84$

So the number of ways $6$ people are selected from $9$ people=$84$
but we have considered two people as one because either will be attending party 

So the actual number of arrangements with two friends not attending party together=$84$$\times$$2$=$168$
So is the Answer $168$ i.e None of these
First off, for the arrangements that do not include the rival, you should see how many parties you can invite from the 8 remaining.  

Then you need to see how many ways you can match up Rival #1 with 5 people from the remaining 8.  Then how many ways can you match up Rival #2 with 5 people from the remaining 8.

ETA: Probably easier method would be ${10\choose 6}-{8\choose 4}$ which is total parties of 6 minus parties containing both rivals.
If neither of the two unamicable friends don't attend the party, we choose $\binom{8}{6} = 28$.

Now if either of them is selected for the party (but not the other), we have $ \binom{2}{1}\cdot \binom{8}{5} = 112$.

Thus the answer will be $112+28= 140$.