Thread: Anyone here good at finite?

04252005, 12:07 PM #1Registered User
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 Nov 2004
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Anyone here good at finite?
I need help with a problem... anyone?

04252005, 12:09 PM #2
I'm good at finite.

04252005, 12:14 PM #3
Im not

04252005, 12:19 PM #4Registered User
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 Nov 2004
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crap, my exam tomorrow is 55% of my mark
I did well during the term, but now im having a mental block

04252005, 12:22 PM #5
Sorry, not a math guy..good luck though.

04252005, 12:23 PM #6
I'm good at math, what's the problem?

04252005, 12:31 PM #7Registered User
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honestly... i think this problem is just dumb... and my brain quit about a week ago, lol
"every customer who orders an extra large pizze today from Pepe's Perfect pizza will be offered the anniversay special: they can either get a small pizza (along with their extra large) for an extra $2, or they can get a second extra large for an extra $5. They cannot get both, an some cumstomers, of course, may not want either. Pepe wants to know how many different possibilities there are for having to provide these special pizzas, assuming that 50 people order extra large pizzas today (i.e 50 people are offered this special). Pepe does not care about which customers order the small of extra large special. He just cares about how many customers order each. (eg. 50 take the small special, or 20 take the small special, 20 take the XL special and 10 take neither.) How many different possibilities are there?
i think this is just a dumb question with too much info....

04252005, 02:17 PM #8Originally Posted by Bliss

04252005, 05:52 PM #9
not much time to work it out to verify, but the only info you need is there are 3 possibilities (x, y, z) and there are 50 customers. so really you have 50, 50, 50. so i BELIEVE its 50! + 50! + 50! which = 9.12422796 × 10^64. check it out and see. Ill work it out and double check it tomorrow morning around 8am and post a definate answer.
max

04252005, 05:53 PM #10
that seems like a lot to me though...but quite possibly could be correct.

04252005, 07:41 PM #11
(0,0,X) x 3 (any variable x,y,z) = 150
(0,X,0) x 3 (any variable x,y,z) = 150 } 0 < x,y,z < 50
(X,0,0) x 3 (any variable x,y,z) = 150
and so on....
(X,Y,0)
(X,0,Y)
(Y,X,0)
(Y,0,X)
(0,X,Y)
(0,Y,X)
this is how i would start it. I think the 10^64 is a little high too, but ya never know...

04252005, 07:55 PM #12Registered User
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 Nov 2004
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hahahahahaha thats for the help guys... I got the answer now
just for ur info.... its a free distribution, all customers are considered identical and there are 3 ways to distribute them, so its (50+31 C 50)= (52C50)

04252005, 08:00 PM #13Originally Posted by max2extreme

04252005, 09:55 PM #14
ok i believe you need to use that formula nCr. The odds of all 50 choosing the extra large are just the same as the 50 people picking any other combination. The answer I got was 125,000, dont ask me to explain it tho
http://mathforum.org/pcmi/hstp/sum20...ing/brian.html

04252005, 10:03 PM #15Registered User
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 Nov 2004
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i already got the answer, its a free distribution problem!!! lol
thanks for trying to help though

04252005, 10:07 PM #16Originally Posted by Bliss

04252005, 10:12 PM #17Registered User
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combination.... so its 52!/50!2!
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