# 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?

• Also consider the scenario when A and B are both not invited. Commented Jun 15, 2022 at 9:15

You can simply say that it is the difference of "all situations" minus "both are invited" : $$C(14,8)-C(12,6)=2079$$

• Is it supposed to be all situations - both are not invited? Commented Jun 15, 2022 at 9:35
• @Abbas If is all situations - both are invited. The situation that is not permissible is that both $A$ and $B$ are invited. Commented Jun 15, 2022 at 9:37
• @The Date Tree: to make your answer clearer you should put quotes and/or parentheses around the phrases "all situations" and "both are invited". Otherwise, it can be hard to parse your statement.
– stux
Commented Jun 15, 2022 at 20:01

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 $$\binom{12}{7} + \binom{12}{7} + \binom{12}{8}$$

• I think this answer could be improved by comparing to OP's method. In particular, by noting that OP's method amounts to counting the first two scenarios, which are distinct, as one, so that OP's number $C(13,8)$ computed $C(12,7)+C(12,8)$ instead of $2 \cdot C(12,7)+C(12,8)$. Commented Jun 15, 2022 at 18:31
• Could you please add the end result as a number? Commented Jun 16, 2022 at 12:32
• @A.L $2\binom{12}{7} + \binom{12}{8} = 2 \cdot 792 + 495 = 2079$. Commented Jun 16, 2022 at 13:09