In my copy of An Introduction to Probability by William Feller (3rd ed, v.1), section I.2(b) begins as follows:

(b) Random placement of r balls in n cells. The more general case of [counting the number of ways to put] $r$ balls in $n$ cells can be studied in the same manner, except that the number of possible arrangements increases rapidly with $r$ and $n$. For $r=4$ balls in $n=3$ cells, the sample space contains already 64 points ...

This statement seems incorrect to me. I think there are $3^4 = 81$ ways to put 4 balls in 3 cells; you have to choose one of the three cells for each of the four balls. Feller's answer of 64 seems to come from $4^3$. It's clear that one of us has made a very simple mistake.

Who's right, me or Feller? I find it hard to believe the third edition of a universally-respected textbook contains such a simple mistake, on page 10 no less. Other possible explanations include:

(1) My copy, a cheap-o international student edition, is prone to such errors and the domestic printings don't contain this mistake.

(2) I'm misunderstanding the problem Feller was examining.

  • 2
    $\begingroup$ $3^4$ looks right. Imagine 4 balls in 1 cell: there is only one possible arrangement. $\endgroup$
    – Henry
    Commented Nov 12, 2011 at 3:12
  • $\begingroup$ Volume $1$ seems to have the same typo, at the Amazon less cheapo price of $129.44$. Found a Canadian price of $189.95 (in Canadian dollars, which are roughly at par with the US dollar). $\endgroup$ Commented Nov 12, 2011 at 3:12
  • $\begingroup$ Are the balls distinguishable? $\endgroup$ Commented Nov 12, 2011 at 3:24
  • $\begingroup$ From the preface to the 1970 revised printing: "In contrast to the first edition, the third was marred by a disturbing number of errata. In the present revised printing, all discovered errata are corrected." However, this revised printing still contains the cited apparent error. (At Amazon, you can "Look Inside" to view this on pp 8-9.) $\endgroup$
    – r.e.s.
    Commented Nov 12, 2011 at 4:02
  • 3
    $\begingroup$ Dimitrije: even not knowing the price you paid, I am sure the content of Feller volume I is worth it. $\endgroup$
    – Did
    Commented Nov 27, 2011 at 12:08

4 Answers 4


It is an error. The answer should is 81. I was really troubled that this book could have gone through 3 editions and a revision reprint and still have this elementary error.

It is easy to see why 81 is the correct number: The first ball can be placed in 3 ways. The second in three ways. By the rs principle, the first two balls can be placed in 3x3 ways, i.e. 9 ways. The third ball can be paced in 3 ways. By the rs principle, the number of ways of placing the first 3 balls is 27. The 4th ball can be place in any one of three ways, hence the total number of ways the balls can be placed is 81.Repeated application of the rs principle gets you to 81, not 64. 64 would be the right answer if we had 4 cells and 3 balls, since 4x4x4 = 64.

What bothers me is that anyone studying on his own will run into these problems with errata everywhere. The first reaction is always: I must be wrong, because these people with their ivy league degrees cannot possibly be wrong!

The guilty secret is that almost all mathematics books have errors of this type. Before you buy any book, first get hold of the errata list. If the error count is too dense, buy a different book.

I hope this helps.


  • 1
    $\begingroup$ And how do you propose to get hold of the errata list? $\endgroup$ Commented Apr 15, 2012 at 23:18

You are right, it is an error.

Amusingly, the 2nd edition contains the same error, but reversed!

For r = 3 balls in n = 4 cells, the sample space contains already 81 points.

It appears that for the 3rd edition he double-corrected this error, leaving the error you have found.


Assume sampling with replacement, there are four possible balls for cell 1, four possible balls for cell 2, and four possible balls for cell 3. So there are 4^3=64 possibilities. (Assuming sampling without replacement, there are four possible balls for cell 1, three for cell 2, and two for cell 3, for a total of 4*3*2=24.) So it seems the original Feller was correct.

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    $\begingroup$ Welcome to MathSE. Use MathJax for math formatting. $\endgroup$
    – Naive
    Commented Dec 17, 2017 at 5:32

I have to go with Feller: each of the 3 balls has one of 4 cell numbers associated with it. That is 4^3. Here are the 64 possibilities 111, 112, 113, 114, 121, 122, 123, 124, 131, 132, 133, 134, 141, 142, 143, 144, 211, 212, 213, 214, 221, 222, 223, 224, 231, 232, 233, 234, 241, 242, 243, 244, 311, 312, 313, 314, 321, 322, 323, 324, 331, 332, 333, 334, 341, 342, 343, 344, 411, 412, 413, 414, 421, 422, 423, 424, 431, 432, 433, 434, 441, 442, 443, 444

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    $\begingroup$ It's $4$ balls in $3$ cells though... $\endgroup$
    – Christoph
    Commented Feb 19, 2014 at 21:23

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