Discussion
Notation :
C(n, r) = n!/(r!*(n-r)!))
n! is factorial n.
|A| cardinality of set A
\(\overline{s}\) is negation of s
The number of ways to place r indistinguishable balls in n(n \(\lt\) r) distinct boxes, with no box left empty, is........... .
Answer: A
The number of ways is C(r - 1, n - 1).
Line the r identical balls up in a row. There are \(r - 1\) gaps between them, and dividing the row into n non-empty groups means choosing \(n - 1\) of those gaps for the dividers. Each choice gives one distribution with no box empty, and every such distribution arises once:
\[ \binom{r-1}{n-1} \]
C(n + r - 1, n - 1) is the stars-and-bars count when boxes may be left empty. C(n + r - 1, r - 1) and C(n - 1, r - 1) swap or misplace the parameters; the last is 0 here, since \(n \lt r\).
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