Question

(sorry for my english)
I have a problem from combinatorics and probability. Assume that we have a bag full of balls. The balls can be divided into "n" cathegories by their weight - there are balls that weigh 1Kg, 2Kg, 3Kg, ..., nKg. Every cathegory contains infinitely many balls and any two from the same cathegory are indistinguishable.

We randomly pick from the bag exactly "k" balls and put them into the pocket. What is the probability of having exactly the set of balls {1Kg, 2Kg, 3Kg, ..., kKg} ( every one weight between 1 and K exactly once) ???

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Answer 1

(Duh, limited number of letters in one post)

I know that the probability equals:
The_number_of_positive_cases / The_number_of_all_cases
and the number of all possible cases, when picking exactly "k" balls, should be:
n ^ k
(because we can pick every of the "n" different balls in every of the "k" picks, so we get that result by using the law of multiplication)
but I have no idea how many positive cases are there.. :-(
Thank you for help.

Answer 2

using the enter key to help parse the information into a readable form would be helpful

Answer 3

Assume that we have a bag full of balls.

The balls can be divided into "n" cathegories by their weight:
1Kg, 2Kg, 3Kg, ..., nKg.

Every category contains infinitely many balls and any two from the same cathegory are indistinguishable.

We randomly pick from the bag exactly "k" balls and put them into the pocket.

What is the probability of having exactly the set of balls
{1Kg, 2Kg, 3Kg, ..., kKg} ( every one weight between 1 and K exactly once) ???

Answer 4

0_o
I used several Enters.. 12 as far as I can see..

Answer 5

12 aint enough :)

Answer 6

still hard to read what its giving us as infformation

Answer 7

Sorry :)
This place is overally not that user friendly as expected.. no BB code etc.. but thats not the point. :-)

Answer 8

k! maybe

Answer 9

k*(k-1)*(k-2)*...*3*2*1 = k!

Answer 10

but then it says we have infinite number; so divide that by infinity? doesnt make sense

Answer 11

Well if they are indistinguishable, then I think it's the same situation as the one when we have only one ball of every weight and after choosing it we put itback into the bag..
So no infinity division needed imo.

Answer 12

the limit as the bottom goes to infinity might be zero :)

Answer 13

ok, and you only have k times to take out stuff

Answer 14

\[\binom{k}{n!(n-k!)}\]maybe?

Answer 15

bad typos

Answer 16

zarkon would know ....

Answer 17

You meant: \[\left(\begin{matrix}n \\ k\end{matrix}\right)\]
?? Wait, will try to think about that.

Answer 18

yeah, good luck with it :)