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Beatricce13
Pe: 2013-05-15T19:37:55+03:00 In: In:

Combinari

1. Stiind ca combinari de n luate cate k=combinari de n luate cate l sa se arate ca k+l=n (k,l,n apartin N , k<= n, l<=n )

2. Aratati ca:

a)k(k-1) combinari de n luate cate k=n(n-1) combinari de n luate cate n
b) 2*1*combinari de n luate cate 2+3*2*combinari de n luate cate 3+…+n(n-1)combinari de n luate cate n
c) 1^2combinari de n luate cate 1+2^2 combinari de n luate cate 2+…+n^2 combinari de n luate cate n

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1 raspuns

  1. gunty
    maestru (V)

    1. Trebuie mentionat ca

        \[ l \ne k \]

    , altfel problema e gresita. Rezolvarea:

        \[ C_n^k = C_n^l \Rightarrow \left\{ \begin{array}{l} k = l \\ k = n - l \Leftrightarrow n = k + l \\ \end{array} \right.. \]

    Deoarece

        \[ l \ne k \]

    , inseamna ca ramane doar cazul in care n=k+l.

    2.
    a)

        \[ k\left( {k - 1} \right)C_n^k = n\left( {n - 1} \right)C_{n - 2}^{k - 2} \Leftrightarrow k\left( {k - 1} \right) \cdot \frac{{n!}}{{\left( {n - k} \right)! \cdot k!}} = n\left( {n - 1} \right) \cdot \frac{{\left( {n - 2} \right)!}}{{\left( {n - k} \right)! \cdot \left( {k - 2} \right)!}} \Leftrightarrow \frac{{n!}}{{\left( {n - k} \right)! \cdot \left( {k - 2} \right)!}} = \frac{{n!}}{{\left( {n - k} \right)! \cdot \left( {k - 2} \right)!}}_{\left( {adev.} \right)} \]

    b)

        \[ 1 \cdot 2 \cdot C_n^2 + 2 \cdot 3 \cdot C_n^3 + ... + \left( {n - 1} \right)n \cdot C_n^n = \sum\limits_{k = 2}^n {\left( {k - 1} \right)k \cdot C_n^k = } \sum\limits_{k = 2}^n {n\left( {n - 1} \right)C_{n - 2}^{k - 2} = } n\left( {n - 1} \right)\sum\limits_{k = 2}^n {C_{n - 2}^{k - 2} } = n\left( {n - 1} \right) \cdot 2^{n - 2} . \]

    c)

        \[ \begin{array}{l} 1^2 C_n^1 + 2^2 C_n^2 + ... + n^2 C_n^n = \sum\limits_{k = 1}^n {k^2 \cdot C_n^k } = n + \sum\limits_{k = 2}^n {k^2 \cdot C_n^k } = n + \sum\limits_{k = 2}^n {\left[ {n\left( {n - 1} \right)C_{n - 2}^{k - 2} + nC_{n - 1}^{k - 1} } \right]} = \\ = n + n\left( {n - 1} \right)\sum\limits_{k = 2}^n {C_{n - 2}^{k - 2} + n\sum\limits_{k = 2}^n {C_{n - 1}^{k - 1} } } = n + n\left( {n - 1} \right)2^{n - 2} + n\left( {2^{n - 1} - 1} \right) = n\left( {n - 1} \right)2^{n - 2} + 2n \cdot 2^{n - 2} = n \cdot 2^{n - 2} \left( {n - 1 + 2} \right) = n\left( {n + 1} \right) \cdot 2^{n - 2} . \\ \end{array} \]

    * Am corectat enuntul si la 2 – a).
    ** S-au folosit formulele:

        \[ \begin{array}{l} k^2 \cdot C_n^k = n\left( {n - 1} \right)C_{n - 2}^{k - 2} + nC_{n - 1}^{k - 1} ,\forall n,k \ge 2. \\ \sum\limits_{k = 0}^n {C_n^k = 2^n } . \\ \end{array} \]

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