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In this section, there were four cases, the 4 combinations of distinguishable and identical balls and cells. I could understand [Distinguishable balls and cells] and [identical balls and distinguishable cells] but I just can not understand this one.

First of all, i don't understand stirling numbers of second kind, i had read a little about one of the types of stirling number while studying circular permutations but I was not very well versed with it (read: i am still confused). Secondly, i don't understand how we 'easily' get the two results they have shown.

I would really appreciate if someone could explain this to me and also a little bit of Sterling numbers.

Thanks!

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    $\begingroup$ You said you understand distinguishable balls and cells? If yes, what happens if the cells are not distinguishable? If there are $r$ non-empty cells, you simply divide by $r!$. That is what Stirling Number of the second kind gives you. $\endgroup$
    – Math Lover
    Dec 16 '20 at 9:08
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    $\begingroup$ If you see the expression given for $S(n, r)$, you see $\frac{1}{r!}$. What you see inside the brackets is what you get using P.I.E. for distinguishable balls and cells. $\endgroup$
    – Math Lover
    Dec 16 '20 at 9:12
  • $\begingroup$ @MathLover oh yes, that was stupid of me, thanks! $\endgroup$ Dec 16 '20 at 9:12
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The set $\{1,2,3\}$

can be partitioned into $\mathbf{three}$ subsets in $\mathbf{one}$ way: $\{\{1\},\{2\},\{3\}\}$ so $S(3,\mathbf{3})=\mathbf{1}$;

into $\mathbf{two}$ subsets in $\mathbf{three}$ ways: $\{\{1,2\},\{3\}\}$, $\{\{1,3\},\{2\}\}$, and $\{\{1\},\{2,3\}\}$, so $S(3,\mathbf{2})=\mathbf{3}$;

and into $\mathbf{one}$ subset in $\mathbf{one}$ way: $\{\{1,2,3\}\}$, $S(3,\mathbf{1})=\mathbf{1}$.

There are $\mathbf{zero}$ ways to partitioned $\{1,2,3\}$ into $\mathbf{four}$ non-empty subsets $S(3,\mathbf{4})=\mathbf{0}.$

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