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x
Joined: 31 Oct 2001
Posts: 1634
Location: Athens, GA |
Math Problem
If you have a pool of 8 friends and can only choose 5 to go camping, how many possible groups could you bring?
What if two friends don't like each other and you couldn't bring them together?
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Mon Mar 10, 2014 7:34 am |
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hassan-i-sabbah
Joined: 10 Nov 2006
Posts: 27424
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69...lol.
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quote:
Originally posted by turtleman
A normal person wouldn't say that in real life because it's ridiculous and insulting. Yet here you are spouting the most hateful garbage that your demons can muster out of your darkened soul. All because of the internet.
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Mon Mar 10, 2014 7:43 am |
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Winchester
Joined: 04 Oct 2011
Posts: 980
Location: Melbourne, Australia |
It's over 9000.
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Mon Mar 10, 2014 8:01 am |
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Bobbo
Joined: 30 Jan 2002
Posts: 1119
Location: Virginia Beach, VA |
36
Possible ways to choose without extra condition is just 8C5 = 8! / (3! 5!) = 56
Let us call the two quarreling friends "the bitches", for ease of notation. These 56 possibilities split up into 3 disjoint cases depending on the number of bitches chosen. Define the following numbers:
a := # of groups with 0 bitches
b := # of groups with 1 bitch
c := # of groups with 2 bitches
Then 56 = a + b + c, and we are just interested in 56 - c, in words the number of choices involving only 1 or 0 bitches so that they don't ruin the whole damn trip. To find c, it's the same as counting the number of ways of selecting any 3 additional people out of the 6 non-bitches, or c = 6C3 = 6! / (3! 3!) = 20. Thus we have 36 options.
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Wartoo
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Mon Mar 10, 2014 8:44 am |
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smurf_king
Joined: 07 Mar 2009
Posts: 4366
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i was deliberating and came to the same, 36.
but then i wondered if this is a trick question, it is certainly not clarified.
the question is wether the possible groups must be of 5, or can be smaller.. say, if so... the possibilities are obviously higher  (not as high as 9000 of course lol)
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http://phoenixtears.ca/
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Mon Mar 10, 2014 8:53 am |
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Jon;
Joined: 13 Oct 2008
Posts: 13966
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"i don't have pet peeves, i have major psychotic fucking hatreds"
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Mon Mar 10, 2014 9:58 am |
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Bobbo
Joined: 30 Jan 2002
Posts: 1119
Location: Virginia Beach, VA |
Bonus question:
Suppose you have 8 friends still. Suppose every person has exactly two enemies which they will not go with, and the hate is mutual. Suppose also that you can't even select 4 or more friends to go without bringing enemies. How many options do you have if you want to bring 3?
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Wartoo
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Mon Mar 10, 2014 10:38 am |
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foonat
Joined: 09 Mar 2003
Posts: 7716
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quote:
Originally posted by smurf_king
i rather be a pedophile than a homo
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Mon Mar 10, 2014 10:56 am |
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Jon;
Joined: 13 Oct 2008
Posts: 13966
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quote:
Originally posted by Bobbo
Bonus question:
Suppose you have 8 friends still. Suppose every person has exactly two enemies which they will not go with, and the hate is mutual. Suppose also that you can't even select 4 or more friends to go without bringing enemies. How many options do you have if you want to bring 3?
why are so many enemies going on a camping trip
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"i don't have pet peeves, i have major psychotic fucking hatreds"
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Mon Mar 10, 2014 12:42 pm |
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Bobbo
Joined: 30 Jan 2002
Posts: 1119
Location: Virginia Beach, VA |
quote:
Originally posted by Jon;
quote:
Originally posted by Bobbo
Bonus question:
Suppose you have 8 friends still. Suppose every person has exactly two enemies which they will not go with, and the hate is mutual. Suppose also that you can't even select 4 or more friends to go without bringing enemies. How many options do you have if you want to bring 3?
why are so many enemies going on a camping trip
Let's say we picked 8 people at random from the war2 forums to go camping, to give a possible real-life application.
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Wartoo
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Mon Mar 10, 2014 1:22 pm |
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x
Joined: 31 Oct 2001
Posts: 1634
Location: Athens, GA |
quote:
Originally posted by Bobbo
36
Possible ways to choose without extra condition is just 8C5 = 8! / (3! 5!) = 56
Let us call the two quarreling friends "the bitches", for ease of notation. These 56 possibilities split up into 3 disjoint cases depending on the number of bitches chosen. Define the following numbers:
a := # of groups with 0 bitches
b := # of groups with 1 bitch
c := # of groups with 2 bitches
Then 56 = a + b + c, and we are just interested in 56 - c, in words the number of choices involving only 1 or 0 bitches so that they don't ruin the whole damn trip. To find c, it's the same as counting the number of ways of selecting any 3 additional people out of the 6 non-bitches, or c = 6C3 = 6! / (3! 3!) = 20. Thus we have 36 options.
Ha, after all these years, I can still rely on the mathematicians in this forum. Thanks. (Same solution as my math book... slightly diff notation.)
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Mon Mar 10, 2014 1:58 pm |
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x
Joined: 31 Oct 2001
Posts: 1634
Location: Athens, GA |
No tricks in the question. I found the original question (without the bitches) to be a peculiar number (56) that can be reached many ways, and was wondering if anyone might reason it differently.
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Mon Mar 10, 2014 2:01 pm |
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x
Joined: 31 Oct 2001
Posts: 1634
Location: Athens, GA |
quote:
Originally posted by Bobbo
Bonus question:
Suppose you have 8 friends still. Suppose every person has exactly two enemies which they will not go with, and the hate is mutual. Suppose also that you can't even select 4 or more friends to go without bringing enemies. How many options do you have if you want to bring 3?
If there are 8 people and each person hates 2 others and the feeling is mutual, wouldn't you need a multiple of 3? I ended up with my last 2 guys simply hating each other because they couldn't select another person to hate, because that person then couldn't hate them back.
allstar (hates blid + chayliss)
blid (hates allstar + chayliss)
chayliss (hates allstar + blid)
dugs (hates equinox + frost)
equinox (hates dugs + frost)
frost (hates dugs + equinox)
ghostnuke (hates hamster)
hamster (hates ghostnuke)
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Mon Mar 10, 2014 2:14 pm |
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Bobbo
Joined: 30 Jan 2002
Posts: 1119
Location: Virginia Beach, VA |
So hating is mutual:
(A hates B) implies (B hates A)
But hating is NOT transitive, meaning:
(A hates B) & (B hates C) does not imply that (A hates C)
Otherwise you are right and we'd need the number of people to be a multiple of 3, so I'm guessing this is what you were thinking?
You'd just need to have "hate loops" of length
at least
3 people. If you made a graph with vertices corresponding to the people and edges corresponding to hate, this just says that every vertex has exactly two edges coming out of it (and at most one edge can join each pair of vertices). Thus we get a disjoint union of circuits or hate loops which have to have at least 3 people in them like you point out. But you could have 4,5,6,7, or 8 people in each one.
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Wartoo
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Mon Mar 10, 2014 2:31 pm |
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Sypher
Joined: 18 Sep 2000
Posts: 5698
Location: Detroit, MI |
Re: Math Problem
quote:
Originally posted by x
If you have a pool of 8 friends and can only choose 5 to go camping, how many possible groups could you bring?
What if two friends don't like each other and you couldn't bring them together?
lol like u really have 8 friends.
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"I tend to thougoughly enjoy my encounters significantly more with 120+ types, as I find them more stimulating. 100-110 people are okay too operating at full capacity." - Paper_Boy
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Mon Mar 10, 2014 10:34 pm |
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TYRYTY
Joined: 09 Oct 2010
Posts: 430
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invest in bitcoins
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Tue Mar 11, 2014 10:29 am |
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Jon;
Joined: 13 Oct 2008
Posts: 13966
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are u interested in topology bobbo
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"i don't have pet peeves, i have major psychotic fucking hatreds"
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Tue Mar 11, 2014 11:31 am |
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Jon;
Joined: 13 Oct 2008
Posts: 13966
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trying to create a worm hole
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"i don't have pet peeves, i have major psychotic fucking hatreds"
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Tue Mar 11, 2014 11:39 am |
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Bobbo
Joined: 30 Jan 2002
Posts: 1119
Location: Virginia Beach, VA |
LOL well I guess topology will help with that. In case you weren't joking, I'd recommend these two books if you want to teach yourself some stuff:
http://www.amazon.com/Topology-2nd-Edition-James-Munkres/dp/0131816292
(best introduction to the subject ever)
http://www.amazon.com/basic-course-algebraic-topology-127/dp/038797430X
(excellent follow-up and gets a little more abstract, into deeper topics of cohomology/homology theories)
Allen Hatcher has a widely-used text which would be a great 3rd step, but his book is not nearly as well-written as the above two (which are perfect for self-teaching). Hatcher's book is a little intense and sloppy at times.
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Wartoo
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Tue Mar 11, 2014 11:47 am |
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Fast Luck
Joined: 11 Oct 2001
Posts: 22805
Location: Penis |
lol
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i zero bagged your mother
quote:
Originally posted by Fast Luck
hassan-i-asher: majorin in takin pictures
dreamin bout wayne from catalina wine mixers
listen little friend stay outta the deep end
cuz you're less street than vampire weekend
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Tue Mar 11, 2014 12:03 pm |
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hassan-i-sabbah
Joined: 10 Nov 2006
Posts: 27424
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Wow, this dog can do everything.
_________________
quote:
Originally posted by turtleman
A normal person wouldn't say that in real life because it's ridiculous and insulting. Yet here you are spouting the most hateful garbage that your demons can muster out of your darkened soul. All because of the internet.
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Tue Mar 11, 2014 12:59 pm |
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7VlesSiah
Joined: 16 Feb 2001
Posts: 2456
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Re: Math Problem
quote:
Originally posted by x
If you have a pool of 8 friends and can only choose 5 to go camping, how many possible groups could you bring?
What if two friends don't like each other and you couldn't bring them together?
8 C 5 = 56
6C4 +6C4 +6C5 = 36
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I have hacks in my brain and I use them.
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Tue Mar 11, 2014 1:38 pm |
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foonat
Joined: 09 Mar 2003
Posts: 7716
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Re: Math Problem
quote:
Originally posted by Shotgun_
quote:
Originally posted by 7VlesSiah
quote:
Originally posted by x
If you have a pool of 8 friends and can only choose 5 to go camping, how many possible groups could you bring?
What if two friends don't like each other and you couldn't bring them together?
8 C 5 = ogrerush
6C4 +6C4 +6C5 = ogrerush
i feel like after so many years of ogre rush jokes, i shouldn't find these as funny as i do
_________________
quote:
Originally posted by smurf_king
i rather be a pedophile than a homo
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Tue Mar 11, 2014 2:41 pm |
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