A man desires to throw a party for some of his friends. In how many ways can he select 8 friends from a group of 14 friends? A man desires to throw a party for some of his friends. In how many ways can he select $8$ friends from a group of $14$ friends if the two of his friends(say ’A’ and ’B’) will not attend the party together?
This is what I've done:
Lets make two groups one for A and one for B
$A$ = $\{A,C,D,E,F,G,H,I,J,K,L,M,N\}$
$B$ = $\{B,C,D,E,F,G,H,I,J,K,L,M,N\}$
Since $A$ and $B$ will not attend together, there is only 13 friends to choose from:
$^{13}C_8 = 1287$ ways to invite.
Is this approach and answer correct?
 A: The method implemented by The Date Tree is optimal.
To understand why your approach is not correct, let's perform a direct count.  There are three possible scenarios:

*

*$A$ is invited, $B$ is not invited, and seven of the other $12$ people are invited.  This can occur in $$\binom{1}{1}\binom{1}{0}\binom{12}{7} = \binom{12}{7}$$ ways.

*$A$ is not invited, $B$ is invited, and seven of the other $12$ people are invited.  This can occur in $$\binom{1}{0}\binom{1}{1}\binom{12}{7} = \binom{12}{7}$$ ways.

*Neither $A$ nor $B$ is invited and eight of the other $12$ people are invited.  This can occur in $$\binom{1}{0}\binom{1}{0}\binom{12}{8} = \binom{12}{8}$$ ways.

Since these three scenarios are mutually exclusive and exhaustive, the number of ways the man can invite eight of his friends to a party if $A$ and $B$ are not both invited is $$\binom{12}{7} + \binom{12}{7} + \binom{12}{8}$$
A: You can simply say that it is the difference of "all situations" minus "both are invited" : $$C(14,8)-C(12,6)=2079$$
