# Friendships and probability [duplicate]

Question 8. Jackie has 14 close friends. Find the number of ways she can invite 5 of them to dinner under the following conditions:

A. There are no restrictions.

B. Two of the friends are married to each other and will not attend separately.

C. Two of the friends are not speaking to each other and will not attend together.

So for no restrictions I think it would just be 14 choose 5.For B, I think it would be 14 choose 5 minus 10 because for each of the 5 "slots" that a friend could take up if one of them is from the married couple and the other isn't there it is a null possibility hence it is subtracted. For C, I think it would be 14 choose 5 minus 5 because there are 5 "slots" and if there are both people present in this set of 5, then that option would be subtracted. Am I going about this correctly? Thanks.

• You have the right beginnings, however you don't seem to have made appropriate calculations for parts b) and c). Look at this more carefully... suppose that two people are married and will only attend together if at all. Among all of the ways that you can invite five to dinner, exactly one of two things will happen. Both of the married couple are invited along with three other friends -or- Neither of the married couple are invited along with three other friends. How many ways can each of these happen? Can you break it down similarly to this for part c)? – JMoravitz Nov 30 '17 at 21:57
• I recommend looking at the Related tab to the right of the screen. There are several very similar problems to this one linked there where the only differences are a bit of flavor and specific numbers, perhaps having 11 friends total instead of 14 and so on, or having a slightly different twist. The methods and approach to a solution will be either identical or able to be modified slightly to suit your specific needs. It is always a good idea when posting a question to check the suggested duplicates or related section before actually posting. – JMoravitz Nov 30 '17 at 22:04

(a) looks okay.

For (b) you can either select neither $2$ of the couple and $5$ of the other $12$ people, or select both people from the couple and $3$ of the other $12$ people giving $${2\choose{0}}\cdot{12\choose{5}}+{2\choose{2}}\cdot{{12\choose{3}}}=1012$$

For (c) we can either select $5$ of the $12$ indifferent people and neither of the friends, or $1$ of the $2$ friends and $4$ of the $12$ indifferent people giving

$${2\choose{0}}\cdot{12\choose{5}}+{2\choose{1}}\cdot{12\choose{4}}=1287$$

I write it out like this because it gives you a check to make sure that the sums of the top and bottoms comes out to be $14$ and $5$ respectively.

• Also, another way to evaluate $c$ is to subtract the prohibited selection from the total. $$\dbinom{14}{5} - \dbinom{2}{2}\cdot\dbinom{12}{3} = 1287$$ – Graham Kemp Dec 1 '17 at 0:15

A) $14\choose5$

B) $12\choose5$+$12\choose3$

C) $12\choose5$+2$12\choose4$

You're right for A.

For B consider 2 states, either the married couple are coming or they're not. If they're not coming then there is 12 choose 5 possibilities.

If they are coming then we have 12 choose 3 other spots.

So the answer is 12 choose 5 plus 12 choose 3.

For C we'll call the 2 friends not speaking A and B. If A is going, then there are 12 choose 4 remaining spots. By symmetry, the same is true for B. If neither is going then there are 12 choose 5 spots.

So the answer is 2 times 12 choose 4 plus 12 choose 5.

• Why 13 choose 4 for C, maybe 12? – user Nov 30 '17 at 22:13