In how many distinct ways can three purple and two pink balls be arranged in a row?
I think you guys can figure it out pretty easily... but I'm too stupid to so help :P!
Six.
P = purple
I = pink
PIPIP
PIPPI
PIIPP
IPPPI
IPPIP
IIPPP
Incidentally, there's an easier way than thinking about every combination. Multiply the two numbers.
What about PPIPI, PIIPP, PPIIP, etc?
deadly got owned.
Incidentally, Ergot, PIIPP was included. Also, PPIIP is simply PIIPP backwards, but I don't know if you want those or not.
I think the formula deadly gave is the "I have X pairs of pants and Y shirts. How many outfits can I make?" formula. Going by that, we assume the days he's wearing shirts he won't wear pants, and vise versa. Sure, Ergot would like that, but..
10, I think? That includes ones that are backwards-versions of other ones:
PPPII
PPIPI
PIPPI
IPPPI
PPIIP
PIPIP
IPPIP
PIIPP
IPIPP
IIPPP
That's a combination or permutation or some shit like that. :(
It's ten. The algebraic way of doing this (combinations) is 5! / (3!2!). There are 5! permutations and you have to divide by 3! because there are 3! arrangements of three purple balls in a row and 2! arrangements of the whatever-color ball in a row and those produce non-distinct permutations.
Quote from: Ender on August 29, 2006, 11:40:13 PM
It's ten. The algebraic way of doing this (combinations) is 5! / (3!2!). There are 5! permutations and you have to divide by 3! because there are 3! arrangements of three purple balls in a row and 2! arrangements of the whatever-color ball in a row and those produce non-distinct permutations.
OMG YOU'RE SO SMART x_x! I never learned that in algebra :(.
Hehe. I forgot everything I learned in pre-calc already. :(
I forgot how to figure premutations. :(
You forgot how to spell them, too.
Either that or he really doesn't know how to recognize the factors preceding a mutation :P