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in this answer it is stated that

For each such choice, the two balls that go into the lucky cell can be chosen in $\binom{n}{2}$ ways

(i.e. the number of ways to choose two balls from $n$ balls to go in one cell is $\binom{n}{2}$)

Then, in the comments it is stated

"Under the most reasonable model, which is that the balls were thrown one at a time towards the cells, independently, with all cells equally likely, it makes no difference whether or not balls and/or cells are distinguishable"

Why under such a model does not not matter whether the balls are indistinguishable or not?

(specifically, I thought that the $\binom{n}{2}$ would be multiplied by $2$ factorial, since for any chosen pair we could reverse the balls and have another choice.)

user106860
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1 Answers1

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If what you consider is how the result will look like, then distinguishable or not matters. While, if what you consider is the entire process of generating the result, at each step you have to make a choice(thrown one at a time towards the cells, independently, with all cells equally likely), even when you pick one from the indistinguishables, as if they're distinguishable(i.e. you have to label them).

linear_combinatori_probabi
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