# A problem on Stirling numbers of second kind. 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!

• 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. Dec 16 '20 at 9:08
• 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. Dec 16 '20 at 9:12
• @MathLover oh yes, that was stupid of me, thanks! Dec 16 '20 at 9:12

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 $$S(3,\mathbf{4})=\mathbf{0}.$$