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I am trying to understand the intuition behind unordered sampling with replacement.

This is the problem I have: I want to distribute $4$ identical balls to $2$ people. Let $1$ represent person $1$, and let $2$ represent person $2$. Then, the potential distributions are $1111, 2222, 1112, 2221, \text{and } 1122$.

Formulaically, we know that there should be ${n+k-1 \choose k}$ distributions. Thus, in this case, there should be $5 \choose 2$ $= 10$ (with $n = 4$ and $k = 2$) distributions.

Where do the other $5$ orderings come from?

snerd8
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neelsnarayan
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  • I agree there are only 4 ways to assign the balls. What leads you to believe you need to use that formula? – snerd8 Feb 24 '24 at 20:40

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Your formula for the number of possible distributions for the given example is incorrect. It's $n+k-1 \choose k-1$ and not $n+k-1 \choose k$. So the correct number of distributions here would be

$${4+2-1 \choose 2-1} = {5 \choose 1} = 5$$

Which is what you rightly figured it should be.

Haris
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  • On this site, it says that the formula is as I had it: https://www.probabilitycourse.com/chapter2/2_1_4_unordered_with_replacement.php – neelsnarayan Feb 25 '24 at 00:38
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    Well, it's an error. If you look at example 2.12 on the same page, they use the correct formula (mentioned in my answer). You can also visit https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)#Theorem_two to be doubly sure. – Haris Feb 25 '24 at 07:15