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# Stirling Numbers

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• #24379
Ashani Dasgupta
Keymaster

#24383

Will try. But need to think about how recursion can play a role here.
Thank You Sir.

#24415
Participant

Sir, solved the first one

#24416
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#24439
swastik pramanik
Participant

Solution for the second problem:

Let us consider $r$ objects which are to be distributed in $n$ identical boxes with no box empty. So,  by definition we can place $r$ objects in $n$ identical boxes in $S(r,n)$.

Let $P=\{p_1,p_2,…,p_r\}$ be the set of $r$ objects and $T=\{t_1,t_2,…,t_n\}$ be the set of $n$ identical boxes.

Now we divide the problem into two disjoint cases.

Case 1:  When box $t_1$ contains only object $p_1$ then,

The remaining $r-1$  objects can be distributed to the remaining $n-1$ boxes in $S(r-1,n-1)$ ways.

Case 2:  When box $t_1$ will have the object $p_1$ and also some other objects then,

The remaining $r-1$ objects are first put in the $n$ boxes in $S(r-1,n)$ ways. After distributing the $r-1$ objects in $n$ boxes each box becomes distinct. Hence, we can place $p_1$ in the $n$ boxes in $n$ ways. Thus, the number of ways of such distribution is $nS(r-1,n)$.

Thus, we have $$S(r,n)=S(r-1,n-1)+nS(r-1,n)$$ as required.

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