# Combinatorics: inviting friends to dinner with and without restrictions

A woman has 14 friends.

(a) Invite 5 friends to dinner without restriction: $$_{14}C_5$$ (b) Invite 5, but two are married and will not attend separately: $$_{14}C_5-(_{14}C_1)-(_{14}C_1)$$ My logic was to take the total number of ways and subtract the number of ways we can invite those two married people separately.

(c) Invite 5, but two are enemies and will always attend separately. $$_{14}C_5-(_{14}C_2)+(_{14}C_1)+(_{14}C_1)$$Similar logic as above: Take the total number of ways, subtract the cases were those two come together, and then add the cases where we invite them separately.

Does my logic make sense?

Your answer to the first question is correct.

In how many ways can a woman invite five of her fourteen friends to dinner if two of those friends are married and will not attend separately?

Method 1: We subtract those cases in which exactly one person from that married couple is selected to attend from the total.

There are $\binom{14}{5}$ ways for the woman to invite five of her fourteen friends to dinner. There are two ways for her to invite exactly one of the two members of the married couple and $\binom{12}{4}$ ways for her to invite four additional friends from the other twelve available friends. Hence, the number of permissible selections is $$\binom{14}{5} - \binom{2}{1}\binom{12}{4}$$

Method 2: The woman either includes the married couple or she does not. If she includes the married couple, she must select three of her other twelve friends as well. If she does not include the married couple, she must select five of her other twelve friends. Hence, the number of permissible selections is $$\binom{2}{2}\binom{12}{3} + \binom{2}{0}\binom{12}{5}$$

In how many ways can a woman invite five of her fourteen friends if two are enemies who will not attend together?

Method 1: The woman invites neither enemy or she invites exactly one of them.

If she invites neither enemy, she must select five of her other twelve friends. If she invites exactly one of them, she must select one of the two enemies and four of her other twelve friends. Hence, the number of permissible selections is $$\binom{2}{0}\binom{12}{5} + \binom{2}{1}\binom{12}{4}$$

Method 2: We subtract the number of selections in which the woman selects both enemies from the total.

If the woman were to select both enemies, she would also have to select three of her other twelve friends. Hence, the number of permissible selections is $$\binom{14}{5} - \binom{2}{2}\binom{12}{3}$$

For b), you can club the friends into one and invite the rest in two ways one to include them and the other to exclude them. Thus the answer is ${12\choose3}+{12\choose5}=1012$. Part c), Subtract cases where they attend together from total ways, so ${14\choose5}-{12\choose3}=1782$.

• For (b), I understand the 12C3, but why is it 12C5, instead of 13C5? We group the couple as 1 so now we're picking 5 from 13 right? – VV6570 Nov 29 '17 at 16:51
• Remember we are excluding the two and hence the remainder is 14-2 = 12 and you are picking the five friends from the 12 remaining. That is the logic – Satish Ramanathan Nov 29 '17 at 16:53