Question

In how many ways can you distribute "a" identical balls to "b" persons?

Answer 1

I believe the answer is:
\[\left(\begin{matrix}a+b-1 \\ b-1\end{matrix}\right)\]
Im working through some examples to make sure...

Answer 2

this is the same question as:
how many solutions does
\[x_1+x_2+\cdots+x_b = a\]
have in integers.

Answer 3

joe is it a permutation or a combination?

Answer 4

i guess you could also write the solution as:
\[\left(\begin{matrix}a+b-1 \\ a\end{matrix}\right)\]
its the same thing.

Answer 5

hmm....I would say its a combination.

Answer 6

b choices for every ball.....b^a ?

Answer 7

i remember a+b-1 and b-1.. its right for sure! But i have a doubt if its a permutation or combination

Answer 8

estudier no.. once you give the ball ur left wid b-1 balls only

Answer 9

a more concrete example would be something like:
how many solutions are there for:
\[x_1+x_2+x_3 = 10\]
if each x has to be a positive integer?
Thats what this question is asking, except the 3 in my example is the b, and 10 is a.

Answer 10

"How many ways can you distribute 10 ball to 3 people?"

Answer 11

those are the same question.

Answer 12

and the answer would be:
\[\left(\begin{matrix}10+3-1 \\ 10\end{matrix}\right) \]
\[\left(\begin{matrix}12 \\ 2\end{matrix}\right) = 66\]

Answer 13

Why does \[b^a\] not work? Explain.

Answer 14

consider a balls with b-1 dividers ...how many ways can you arrange

\[{a+b-1\choose b-1}\]

Answer 15

lets use concrete examples. How many ways can you distribute 5 balls to 2 people?
0 5
1 4
2 3
3 2
4 1
5 0
There are 6 ways.

Answer 16

yeah, zarkon gots it :)

Answer 17

how do u get \[\left(\begin{matrix}a+b-1 \\ b-1\end{matrix}\right)\]

Answer 18

looking at my example again (5 balls and 2 people)
this is the same as:
How many positive integer solutions are there to
\[x_1+x_2 = 5\]
you split the 5 into ones:
1 1 1 1 1
and you are going to put a plus somewhere in there.
So you are asking, "how many ways can I rearrange 6 things, where 5 of them are identical?"
six comes from 5+2-1

Answer 19

So with a balls and b people, you would split the a into ones (there will be a of them), and you are going to put b-1 plus signs somewhere in them. So you are asking "how many ways can i rearrange a+b-1 things, where a of them are identical, and b-1 are identical?"

Answer 20

sry if that doesnt make sense >.<

Answer 21

thanx everyone..:)

Answer 22

i like to look at it like this

xx|xxxx|xxx|x
here we have 10x's (balls) and 3 dividers (which makes 4 boxes)

there are 10+4-1=10+3=13 objects here so 13! number of ways to arrange them. but permuting the x's and the |'s in place does nothing so I need to get rid of those permnutations

10! ways to arrange the x's and 3! ways to arrange the |'s

so

\[\frac{13!}{10!3!}=\frac{(10+4-1)!}{10!(4-1)!}={10+4-1\choose 4-1}\]

Answer 23

I remember when my professor asked us to come up with the formula <.< i came upon it by looking at a million cases <.< then he asked me to prove it and i tried a double induction >.> then he showed me what zarkon and i posted. I wanted to cry lol

Answer 24

thanx for the medal :)

Answer 25

Ah, I see I have everything distinguishable...duh!