Wait, I think I might have it. Do you do 8 choose 5, and then 6 choose 4 and 2 choose 1 and subtract the latter from the former (for the first part of the problem)?
Hello!
I'm doing the question 1.10 from the Ross text and can't figure out how to get the answer they provide. The question is:
"A woman has 8 friends, of whom she will invite 5 to a tea party. How many choices has she if 2 of the friends are feuding and will not attend together? How many choices has she if 2 of her friends will only attend together?"
I know that to select 5 out of the 8, it's 8 choose 5 or (8*7*6*5*4)/(5*4*3*2*1). But I can't figure out how to make sure that friends A and B are not together. I know that it's not necessarily the case that these women will even be part of the 5. Can you suggest the next step?
Thanks!
The questions are together in the text A First Course in Probability, Sheldon Ross.
For the second question, we could choose 5 from the 8 that do not include those 2 people (so 6 choose 5), or we could choose 3 of the 6 and then 2 of the 2 remaining. That would be (6*5*4)/(3*2*1) + (6*5*4*3*2)/(5*4*3*2*1). Right?
Can you help me get to the logic behind that method? I've gotten to the right number answers, now, but I want to understand the most simple and logical way to get to the solution.
for 8 choose 5 minus 6 choose 3, it looks to me that you take the selection of 5 from 8 and then take out the selection of 3 from 6, because the remainder would be taking 2 out of the 2 feuding friends.
And for the other, it's 6 choose 5 + 6 choose 3, because you could either choose 5 out of the 6 non friends, or you could select 3 from the non-friends and 2 of the friends, right?
(Also, I would have thought this belonged in probability rather than a university-level forum! Sorry about that.)