Counting $k$-tuples from non-distinct collection of elements.












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Assume that we have a collection of objects ${n_1,n_1,n_2,n_2,...,n_k,n_k}$. In how many ways can we construct a set ${n_1,...,n_k}$ consisting of distinct $k$-tuples?






combinatorics







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    0












    $begingroup$


    Assume that we have a collection of objects ${n_1,n_1,n_2,n_2,...,n_k,n_k}$. In how many ways can we construct a set ${n_1,...,n_k}$ consisting of distinct $k$-tuples?






    combinatorics







    share|cite|improve this question









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      0












      0








      0





      $begingroup$


      Assume that we have a collection of objects ${n_1,n_1,n_2,n_2,...,n_k,n_k}$. In how many ways can we construct a set ${n_1,...,n_k}$ consisting of distinct $k$-tuples?






      combinatorics







      share|cite|improve this question









      $endgroup$




      Assume that we have a collection of objects ${n_1,n_1,n_2,n_2,...,n_k,n_k}$. In how many ways can we construct a set ${n_1,...,n_k}$ consisting of distinct $k$-tuples?






      combinatorics




      combinatorics






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      asked Dec 7 '18 at 8:01









      user430191user430191

      1799




      1799






















          1 Answer
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          $begingroup$

          You must choose one from each two-element multiset ${n_i, n_i}$, and you must choose $k$ times. Thus there are $2^k$ ways in total.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you very much for the answer. Thus similarly, if we want for instance a collection where two of the $n_i$'s are equal only, then we have $2^{k-1}$ ways right?
            $endgroup$
            – user430191
            Dec 7 '18 at 8:14










          • $begingroup$
            I don’t quite understand what you mean. If only two $n_i$’s are equal there are only two ways to select. If only two are non-equal then we would have $2^{k-1}$ ways.
            $endgroup$
            – William Sun
            Dec 7 '18 at 8:19










          • $begingroup$
            I mean the number of ways, a collection like ${n_1,...,n_j,n_j,....,n_k}$ where all but the $n_j$'s are distinct.
            $endgroup$
            – user430191
            Dec 7 '18 at 8:21










          • $begingroup$
            @user430191 Then clearly there are only two ways, since only $n_j$ can alter.
            $endgroup$
            – William Sun
            Dec 7 '18 at 8:23










          • $begingroup$
            Perhaps I don't phrase that correctly; say that we choose from the beginning that both $n_j$'s are in the collection and we want all the other $k-2$ elements being distinct. In how many ways can we construct such a set this time?
            $endgroup$
            – user430191
            Dec 7 '18 at 8:28











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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          You must choose one from each two-element multiset ${n_i, n_i}$, and you must choose $k$ times. Thus there are $2^k$ ways in total.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you very much for the answer. Thus similarly, if we want for instance a collection where two of the $n_i$'s are equal only, then we have $2^{k-1}$ ways right?
            $endgroup$
            – user430191
            Dec 7 '18 at 8:14










          • $begingroup$
            I don’t quite understand what you mean. If only two $n_i$’s are equal there are only two ways to select. If only two are non-equal then we would have $2^{k-1}$ ways.
            $endgroup$
            – William Sun
            Dec 7 '18 at 8:19










          • $begingroup$
            I mean the number of ways, a collection like ${n_1,...,n_j,n_j,....,n_k}$ where all but the $n_j$'s are distinct.
            $endgroup$
            – user430191
            Dec 7 '18 at 8:21










          • $begingroup$
            @user430191 Then clearly there are only two ways, since only $n_j$ can alter.
            $endgroup$
            – William Sun
            Dec 7 '18 at 8:23










          • $begingroup$
            Perhaps I don't phrase that correctly; say that we choose from the beginning that both $n_j$'s are in the collection and we want all the other $k-2$ elements being distinct. In how many ways can we construct such a set this time?
            $endgroup$
            – user430191
            Dec 7 '18 at 8:28
















          1












          $begingroup$

          You must choose one from each two-element multiset ${n_i, n_i}$, and you must choose $k$ times. Thus there are $2^k$ ways in total.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you very much for the answer. Thus similarly, if we want for instance a collection where two of the $n_i$'s are equal only, then we have $2^{k-1}$ ways right?
            $endgroup$
            – user430191
            Dec 7 '18 at 8:14










          • $begingroup$
            I don’t quite understand what you mean. If only two $n_i$’s are equal there are only two ways to select. If only two are non-equal then we would have $2^{k-1}$ ways.
            $endgroup$
            – William Sun
            Dec 7 '18 at 8:19










          • $begingroup$
            I mean the number of ways, a collection like ${n_1,...,n_j,n_j,....,n_k}$ where all but the $n_j$'s are distinct.
            $endgroup$
            – user430191
            Dec 7 '18 at 8:21










          • $begingroup$
            @user430191 Then clearly there are only two ways, since only $n_j$ can alter.
            $endgroup$
            – William Sun
            Dec 7 '18 at 8:23










          • $begingroup$
            Perhaps I don't phrase that correctly; say that we choose from the beginning that both $n_j$'s are in the collection and we want all the other $k-2$ elements being distinct. In how many ways can we construct such a set this time?
            $endgroup$
            – user430191
            Dec 7 '18 at 8:28














          1












          1








          1





          $begingroup$

          You must choose one from each two-element multiset ${n_i, n_i}$, and you must choose $k$ times. Thus there are $2^k$ ways in total.






          share|cite|improve this answer









          $endgroup$



          You must choose one from each two-element multiset ${n_i, n_i}$, and you must choose $k$ times. Thus there are $2^k$ ways in total.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 7 '18 at 8:06









          William SunWilliam Sun

          471111




          471111












          • $begingroup$
            Thank you very much for the answer. Thus similarly, if we want for instance a collection where two of the $n_i$'s are equal only, then we have $2^{k-1}$ ways right?
            $endgroup$
            – user430191
            Dec 7 '18 at 8:14










          • $begingroup$
            I don’t quite understand what you mean. If only two $n_i$’s are equal there are only two ways to select. If only two are non-equal then we would have $2^{k-1}$ ways.
            $endgroup$
            – William Sun
            Dec 7 '18 at 8:19










          • $begingroup$
            I mean the number of ways, a collection like ${n_1,...,n_j,n_j,....,n_k}$ where all but the $n_j$'s are distinct.
            $endgroup$
            – user430191
            Dec 7 '18 at 8:21










          • $begingroup$
            @user430191 Then clearly there are only two ways, since only $n_j$ can alter.
            $endgroup$
            – William Sun
            Dec 7 '18 at 8:23










          • $begingroup$
            Perhaps I don't phrase that correctly; say that we choose from the beginning that both $n_j$'s are in the collection and we want all the other $k-2$ elements being distinct. In how many ways can we construct such a set this time?
            $endgroup$
            – user430191
            Dec 7 '18 at 8:28


















          • $begingroup$
            Thank you very much for the answer. Thus similarly, if we want for instance a collection where two of the $n_i$'s are equal only, then we have $2^{k-1}$ ways right?
            $endgroup$
            – user430191
            Dec 7 '18 at 8:14










          • $begingroup$
            I don’t quite understand what you mean. If only two $n_i$’s are equal there are only two ways to select. If only two are non-equal then we would have $2^{k-1}$ ways.
            $endgroup$
            – William Sun
            Dec 7 '18 at 8:19










          • $begingroup$
            I mean the number of ways, a collection like ${n_1,...,n_j,n_j,....,n_k}$ where all but the $n_j$'s are distinct.
            $endgroup$
            – user430191
            Dec 7 '18 at 8:21










          • $begingroup$
            @user430191 Then clearly there are only two ways, since only $n_j$ can alter.
            $endgroup$
            – William Sun
            Dec 7 '18 at 8:23










          • $begingroup$
            Perhaps I don't phrase that correctly; say that we choose from the beginning that both $n_j$'s are in the collection and we want all the other $k-2$ elements being distinct. In how many ways can we construct such a set this time?
            $endgroup$
            – user430191
            Dec 7 '18 at 8:28
















          $begingroup$
          Thank you very much for the answer. Thus similarly, if we want for instance a collection where two of the $n_i$'s are equal only, then we have $2^{k-1}$ ways right?
          $endgroup$
          – user430191
          Dec 7 '18 at 8:14




          $begingroup$
          Thank you very much for the answer. Thus similarly, if we want for instance a collection where two of the $n_i$'s are equal only, then we have $2^{k-1}$ ways right?
          $endgroup$
          – user430191
          Dec 7 '18 at 8:14












          $begingroup$
          I don’t quite understand what you mean. If only two $n_i$’s are equal there are only two ways to select. If only two are non-equal then we would have $2^{k-1}$ ways.
          $endgroup$
          – William Sun
          Dec 7 '18 at 8:19




          $begingroup$
          I don’t quite understand what you mean. If only two $n_i$’s are equal there are only two ways to select. If only two are non-equal then we would have $2^{k-1}$ ways.
          $endgroup$
          – William Sun
          Dec 7 '18 at 8:19












          $begingroup$
          I mean the number of ways, a collection like ${n_1,...,n_j,n_j,....,n_k}$ where all but the $n_j$'s are distinct.
          $endgroup$
          – user430191
          Dec 7 '18 at 8:21




          $begingroup$
          I mean the number of ways, a collection like ${n_1,...,n_j,n_j,....,n_k}$ where all but the $n_j$'s are distinct.
          $endgroup$
          – user430191
          Dec 7 '18 at 8:21












          $begingroup$
          @user430191 Then clearly there are only two ways, since only $n_j$ can alter.
          $endgroup$
          – William Sun
          Dec 7 '18 at 8:23




          $begingroup$
          @user430191 Then clearly there are only two ways, since only $n_j$ can alter.
          $endgroup$
          – William Sun
          Dec 7 '18 at 8:23












          $begingroup$
          Perhaps I don't phrase that correctly; say that we choose from the beginning that both $n_j$'s are in the collection and we want all the other $k-2$ elements being distinct. In how many ways can we construct such a set this time?
          $endgroup$
          – user430191
          Dec 7 '18 at 8:28




          $begingroup$
          Perhaps I don't phrase that correctly; say that we choose from the beginning that both $n_j$'s are in the collection and we want all the other $k-2$ elements being distinct. In how many ways can we construct such a set this time?
          $endgroup$
          – user430191
          Dec 7 '18 at 8:28


















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