Feller Probability - putting r balls in n cells This is fundamental, but I think I should have it clarified. In the first chapter of Feller's Introduction to probability theory, the author writes

I always thought that if we label the cells $1,2,3,\ldots,n$ and form $r$-tuples of the form $\{1,1,1\}$,$\{1,1,2\}$,...,$\{4,4,4\}$, there are $n^r=4^3=64$ sample points. Please correct me.


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*Quasar.

 A: This amounts to the number of functions $f:A\to B$ from a set $A$ with $3$ elements (the three balls) $a_1$, $a_2$ and $a_3$, say, to a set $B$ with $4$ elements (the four cells). Indeed, you must assign one of the four cells to each of the three balls. Let's count:


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*Make a choice for $f(a_1)$. You have $4$ options.

*Make a choice for $f(a_2)$. You have $4$ options.

*Make a choice for $f(a_3)$. You have $4$ options.


All in all, the Rule of product leads to $4\cdot4\cdot4=4^3=64$ points, so you are correct and the book is wrong.
Your (equivalent) approach is also correct. You are calculating $|X_1\times X_2\times X_3|$ where $|X_i|=4$ ($i=1,2,3)$. The first component gives the cell of one (fixed) ball, the second component gives the cell of one of the remaining two balls and the third component gives the cell you assign to the last (fixed) ball.
A: I checked into my book, and it clearly says $4$ balls and $3$ cells and $64$ points. I saw your version in an online pdf. I think it is wrong in both versions. ☺ 

