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• #24379
Ashani Dasgupta
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#24383
Ananya Bhattacharya
Participant

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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