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  • #24379
    Ashani Dasgupta
    Keymaster

    Stirling Numbers of Second Kind

    Stirling number recursion

    #24383
    Ananya Bhattacharya
    Participant

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

    #24415
    Prasad s
    Participant

    Sir, solved the first one

    #24416
    Prasad s
    Participant

    uploading again

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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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