Anyone here good at finite?

[Bliss]

I need help with a problem... anyone?

[Polska]

I'm good at finite.

[SplinterCell]

Im not

[Bliss]

crap, my exam tomorrow is 55% of my mark
I did well during the term, but now im having a mental block

[Juddman]

Sorry, not a math guy..good luck though.

[fitnessguy]

I'm good at math, what's the problem?

[Bliss]

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....

[fitnessguy]

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....

OK I was thinking or hoping you were talking maybe a finite integral problem, or a limit problem. Statistics and probability aren't my strongpoints I'm sorry. I'm good at Calc I-III, Trig, Alg, and Physics. sorry bout that. i would have to work it out a long way with X, Y, & Z....i'm sure there is a much shorter way.

[max2extreme]

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

[max2extreme]

that seems like a lot to me though...but quite possibly could be correct.

[fitnessguy]

(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...

[Bliss]

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+3-1 C 50)= (52C50)

[JDawg1536]

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

I havent taken math in a few years, but i remember doing this stuff. Im pretty sure that number is off. cuz think about it logically. if there were 2 choices how many possibilities would there be? 50 right? so to add one more possibility wouldnt increase the number that much. ill remember how to do it in a few minutes.........

[JDawg1536]

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

[Bliss]

i already got the answer, its a free distribution problem!!! lol
thanks for trying to help though

[fitnessguy]

i already got the answer, its a free distribution problem!!! lol
thanks for trying to help though

i don't understand your answer... (52C50) what is the C? a constant?

[Bliss]

combination.... so its 52!/50!2!