Puzzler for you guys

[gopunk]

if you have 10 frogs, 10 bunnies, 10 cats, 40 numbered cages, how many different ways can you place the animals into the cages? the cages are distinguishable, animals are not distinguisable from animals of the same type, and there are no restrictions as to type or number of animals in each cage.

 

[Evadman]

10! 9! 8! then maybe 10! for the extra 10 cages?

 

[gopunk]

btw i have no idea either, i just came back from a final and this was the one subproblem i couldn't get i got the extra credit though, so hopefully i'll get partial credit and still get a full score

 

[DaveSimmons]

discrete mathematics -- ow! brain hurts!! and I was here in OT to escape from rational thought

too much like work for me, but a good puzzle

 

[Viper GTS]

40 cages
3 animal types
30 potential slots for animals, each of which can be occupied by either an animal or nothing

I say 3600, but I'm probably wrong.

Viper GTS

 

[gopunk]

Originally posted by: Viper GTS
40 cages
3 animal types
30 potential slots for animals, each of which can be occupied by either an animal or nothing

I say 3600, but I'm probably wrong.

Viper GTS

my brain is fried... why 30 potential slots?

 

[gopunk]

Originally posted by: DaveSimmons
discrete mathematics -- ow! brain hurts!! and I was here in OT to escape from rational thought

too much like work for me, but a good puzzle

hehe yea my brain hurts too... ugh dumb finals. one more on monday, then i'm done! whoopee! i'm celebrating by having my wisdom teeth pulled out on tuesday

 

[cressida]

use peigon hole

 

[gopunk]

Originally posted by: SoloKid
use peigon hole

eh how?

 

[rob3rt]

If the cages are distinguishable, and the animals aren't, and you also don't care if the animals are mixed in with each other, then...i think you want to solve the # of solutions to this problem

x1+x2+x3+ . . . + x40 = 30

the answer is (30+40-1) choose (40-1) = 69 C 39 = BIG!!

hmm now that i think of it, i could be wrong since i didn't account for the 3 types of animals, but then again you don't care if you mix em in...HMMMM

Edit: sorry meant to say distinguishable cages

 

[Swag1138]

The animals are all indistinguishable from each other, so all you have to do is 10! three times.....or 40! down 30....something like that.....

brain....hurt.......need......caffeine.....

 

[Swag1138]

Originally posted by: rob3rt
If the cages are indistinguishable, and the animals aren't, and you also don't care if the animals are mixed in with each other, then...i think you want to solve the # of solutions to this problem

x1+x2+x3+ . . . + x40 = 30

the answer is (30+40-1) choose (40-1) = 69 C 39 = BIG!!

hmm now that i think of it, i could be wrong since i didn't account for the 3 types of animals, but then again you don't care if you mix em in...HMMMM

I think it was one animal per cage, and the cages are numbered, so they have numbers on them 1-40. Its the animals that are indistinguishable from animals of their own type, so what you have to take into account in this problem is 3 types of 10 animals each, fit into 40 cages.

 

[Cyberian]

Originally posted by: rob3rt
If the cages are indistinguishable, and the animals aren't, and you also don't care if the animals are mixed in with each other, then...i think you want to solve the # of solutions to this problem

I thought that the cages were numbered.