31 people are in a room. they each shake hands with one another

clevohardcore - 6-30-2009 at 05:14 AM

How many handshakes are given?

XHonusWagnerX - 6-30-2009 at 06:40 AM

930?

Dave - 6-30-2009 at 06:51 AM

ummm 961?

XHonusWagnerX - 6-30-2009 at 07:28 AM

Quote:

Originally posted by Dave
ummm 961?

I think that the 'trick/twist' to the question is if YOU'RE one of the 31 people and everyone shakes hands with each other than you dont shake hands with your self.

So technically its 31 people and each one shakes hands with 30 people.

But I could be totally wrong!

moforn - 6-30-2009 at 08:57 AM

hmmm... remember not to count the same handshake twice- I'm too tired to add it up right now but I'm pretty sure the algorithim would be 30+29+28+27...etc.
The first guy would shake hands with 30 guys, but then you wouldn't count the second guys handshake with the first again so he only gets 29, and so on down the line.

XHonusWagnerX - 6-30-2009 at 10:10 AM

Quote:

Originally posted by moforn
hmmm... remember not to count the same handshake twice- I'm too tired to add it up right now but I'm pretty sure the algorithim would be 30+29+28+27...etc.
The first guy would shake hands with 30 guys, but then you wouldn't count the second guys handshake with the first again so he only gets 29, and so on down the line.

damn thats right....

Hmmm..... 465?

Dave - 6-30-2009 at 10:29 AM

Look at that Lionel your not just a dirty hairly little man, you have some reasoning Skrills too.

clevohardcore - 6-30-2009 at 10:31 AM

Yep. GREAT JOB!

There is a algebra formula too. I think its n(n-1)/2. RANDOM would know that one.

upyerbum - 6-30-2009 at 10:45 AM

Lionel said "algorithim."

Discipline - 6-30-2009 at 02:30 PM

Quote:

Originally posted by Dave
Look at that Lionel your not just a dirty hairly little man, you have some reasoning Skrills too.

Yeah, he's got SKRILLS!!!!! :tumble:

random - 6-30-2009 at 03:53 PM

I didn't remember the formula, but the addition gets pretty easy with 30 + 29 + 28 + ... + 1 (and I suck at arithmetic). You can pair up the first and last numbers, and you can combine them.

30 + 1 = 31
29 + 2 = 31
28 + 3 = 31
. .
. .
. .
16 + 15 = 31

So you just get 15 pairs that add up to 31, and 31*15=465.

clevohardcore - 6-30-2009 at 09:08 PM

n(n-1)/2 I believe is the equation. Random that example is awesome though. Crazy how those numbers come out like that every time. Pairing them up and shit.

random - 6-30-2009 at 09:19 PM

clevo,

not as amazing as it seems... all you're doing is subtracting 1 from the first number and adding 1 to the second number.

i think you're right about the equation though. the formula should work with an odd number or terms, too, but my way means you have to be careful.. like if there were 32 people, you'd do 31+1 and so on, but then you'd get stuck with 15 by itself.

holy fuck, i'm a nerd.

random - 6-30-2009 at 09:21 PM

oh, and clevo, there's a similar formula for quadratic equations that's much cooler... you hate FOIL? the equation takes care of it for you. it's a neat thing where you create this "tree" and just combine collected "branches". if you're interested, i'll find you a link... it really is one of those things that fascinating in a nerdy way.

clevohardcore - 6-30-2009 at 11:11 PM

My final math class is going on right now. It's all about things like this. Its a education math clas and its how to do things as a young child or how to figure problems out a different way than normal It's hard as nails for me of course, but you'd have a blast in it I'm sure. Deals with a lot of methods and stuff.

Dave - 6-30-2009 at 11:19 PM

this has to be the geekyist post ever on this board.......

defstarsteve - 7-1-2009 at 12:41 AM

okay lets get nerdy
what is formual for figuring out how many possible 3 number combos on a standard combination lock,
let's say from 0 to 31

and we can do it both ways with numbers being able to repeat and without

repeating
000
001
to 31 31 31

not repeating
and 012
013
014

how many combintaions are possible for both
I did this one saturday morning when I was 12 for the hell of it...
damn if I remember how or remember the answers.

random - 7-1-2009 at 05:01 AM

Quote:

Originally posted by defstarsteve
okay lets get nerdy
what is formual for figuring out how many possible 3 number combos on a standard combination lock,
let's say from 0 to 31

and we can do it both ways with numbers being able to repeat and without

repeating
000
001
to 31 31 31

not repeating
and 012
013
014

how many combintaions are possible for both
I did this one saturday morning when I was 12 for the hell of it...
damn if I remember how or remember the answers.

Repeating: 32*32*32
Not repeating: 32*31*30

random - 7-1-2009 at 05:24 AM

Oh, and the thing I was talking about before is Pascal's Triangle. http://en.wikipedia.org/wiki/Pascal's_triangle

To the right of the "Contents", wikipedia has a neat little illustration of how it's constructed, so you never have to remember the terms. But it's a quick and easy way to write out the expansion of anything of the form (x+y)^n. The triangle gives the coefficients on each term.

(x+y)^0 = 1
(x+y)^1 = x + y
(x+y)^2 = x^2 + 2xy + y^2
(x+y)^3 = x^3 + 3(x^2)y + 3x(y^2) + y^3

Compare that with the first four lines of Pascal's triangle...

1
/ \
1 1
/\ / \
1 2 1
/ \ /\ / \
1 3 3 1

To get any number, just add up the numbers that have lines going down to it... so the next line would be...

1 4 6 4 1

random - 7-1-2009 at 05:25 AM

damn, i just spent time making that triangle look nice, and all my extra spaces were deleted to make it look like dog shit

Enyo - 7-1-2009 at 10:20 AM

random- your brain is amazing. Math has always been impossible for me.

DaveMoral - 7-1-2009 at 11:42 AM

I like that there's open nerdery on this forum.

clevohardcore - 7-1-2009 at 12:20 PM

I wish I was a nerd. Math is not my thing. :sniff:

random - 7-1-2009 at 01:33 PM

hey clevo, i got the package from BD today. thanks!!! you didn't need to do it, but i definitely appreciate it.

Jason the Magnificent - 7-1-2009 at 02:24 PM

"what if c-a-t really spelled dog?"
-Ogre

MyOwnWay - 7-4-2009 at 08:52 AM

Quote:

Originally posted by random

(x+y)^0 = 1
(x+y)^1 = x + y
(x+y)^2 = x^2 + 2xy + y^2
(x+y)^3 = x^3 + 3(x^2)y + 3x(y^2) + y^3

Compare that with the first four lines of Pascal's triangle...

1
/ \
1 1
/\ / \
1 2 1
/ \ /\ / \
1 3 3 1

To get any number, just add up the numbers that have lines going down to it... so the next line would be...

1 4 6 4 1

I'll be honest... This is something I never imagined to see on a Hardcore messageboard... Ever.

XHonusWagnerX - 7-4-2009 at 10:56 AM

this thread gives me a headache and makes me feel dmbu

Spoiler - 7-4-2009 at 11:16 AM

Quote:

Originally posted by MyOwnWay

Quote:

I'll be honest... This is something I never imagined to see on a Hardcore messageboard... Ever.

my brain hurts