1. ## sampling problem

hello
the problem is that suppose i have n distinct balls(or anything) in a box..i have to take r samples ...i.e i have to pick r balls out of those n one by one ...now there are four cases
1.i take the the mth ball without replacing the previous m-1 balls and the sequence in which i take the balls matters
2.same as one but here the sequence doesnot matter
3. i replace the previously taken balls before picking out the next one and the sequence in which they come out matters
4.same as 3 but the sequence doesnot matter

problem is that in each case in how many possible ways the balls can be organized ...or we can say that in each case how many elements are there in sample space ....

now first three cases i have a general solution for them and i understand them ...but in case of 4th i know the answer but dont know how it came up
..the ans is
$
\binom{n+r-1}{r}
$

2. $\binom{n+r-1}{r}$ is the number of ways to select $r$ items from a variety of $n$ different kinds (each kind has at least $r$ available items).

3. i dont understand wat u r trying to say and how it proves the 4th ;part...
$
\binom{n+r-1}{r}
$

this means that we have total of n+r-1 objects and we have to choose r out of them and the sequence doesnot matter to us ...

waiting for a clarification
thanx

4. Originally Posted by corleone2463
i dont understand wat u r trying to say and how it proves the 4th ;part...
$
\binom{n+r-1}{r}
$
Suppose that you go into an ice cream shop which offers twenty-one flavors of ice cream.
They have a special on, a plate of five scoops of any and all flavors.
For example: I would order two scoops of chocolate, two scoops of vanilla and one of strawberry.
How many different selections are possible?
Answer: $\binom{5+21-1}{5}$
This is known as multi-selections $\binom{K+N-1}{K}$ select K items from N difference possibilities.

5. yeah i know that but im asking a proof of that the no of possibilities actualy equals to that factorial

6. Originally Posted by corleone2463
but im asking a proof of that the no of possibilities actualy equals to that factorial
Ok, here is the proof.
If we want to know the number of way to put K identical objects into N different cells that can be modeled using K o’s and N-1 ‘|’.
For example: put five identical balls into three named cells.
This model $o|ooo|o$ says put one ball in the A cell, three in the B cell and one in the C cell.

This model $oooo||o$ says put four balls in the A cell, none in the B cell and one in the C cell.

So any rearrangement of the string $ooooo||$ represents one way to put those five identical balls into three different cells.
The number of ways to rearrange that string is $\frac{7!}{(5!)(2!)}=\binom{5+3-1}{5}$.

To generalize, the number of ways to put K identical objects into N different cells is
$\binom{K+N-1}{K}$.

7. thankyou very much ...this is a good and simple proof
can you suggest some such interesting problems like this one to solve in probability