Counting sequences with exactly d distinct elements












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What is the number of sequences $(x_1,...,x_q) in mathcal{X}^q$ with exactly $d$ distinct elements, for some $1 leq d leq q$?






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    What is the number of sequences $(x_1,...,x_q) in mathcal{X}^q$ with exactly $d$ distinct elements, for some $1 leq d leq q$?






    combinatorics







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      $begingroup$


      What is the number of sequences $(x_1,...,x_q) in mathcal{X}^q$ with exactly $d$ distinct elements, for some $1 leq d leq q$?






      combinatorics







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      What is the number of sequences $(x_1,...,x_q) in mathcal{X}^q$ with exactly $d$ distinct elements, for some $1 leq d leq q$?






      combinatorics




      combinatorics






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      asked Jan 24 at 6:29









      Alex GriloAlex Grilo

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          You don't say what $X$ is, but if it has $x$ elements, then the answer is
          $binom{x}d$ times the solution when $X={1,2,ldots,d}$. In that case, the solution is the number of surjections from ${1,2,ldots,q}$
          to ${1,2,ldots,d}$ which is $d!S(q,d)$ with $S(q,d)$ a Stirling number of
          the second kind.






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            $begingroup$

            You don't say what $X$ is, but if it has $x$ elements, then the answer is
            $binom{x}d$ times the solution when $X={1,2,ldots,d}$. In that case, the solution is the number of surjections from ${1,2,ldots,q}$
            to ${1,2,ldots,d}$ which is $d!S(q,d)$ with $S(q,d)$ a Stirling number of
            the second kind.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              You don't say what $X$ is, but if it has $x$ elements, then the answer is
              $binom{x}d$ times the solution when $X={1,2,ldots,d}$. In that case, the solution is the number of surjections from ${1,2,ldots,q}$
              to ${1,2,ldots,d}$ which is $d!S(q,d)$ with $S(q,d)$ a Stirling number of
              the second kind.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                You don't say what $X$ is, but if it has $x$ elements, then the answer is
                $binom{x}d$ times the solution when $X={1,2,ldots,d}$. In that case, the solution is the number of surjections from ${1,2,ldots,q}$
                to ${1,2,ldots,d}$ which is $d!S(q,d)$ with $S(q,d)$ a Stirling number of
                the second kind.






                share|cite|improve this answer









                $endgroup$



                You don't say what $X$ is, but if it has $x$ elements, then the answer is
                $binom{x}d$ times the solution when $X={1,2,ldots,d}$. In that case, the solution is the number of surjections from ${1,2,ldots,q}$
                to ${1,2,ldots,d}$ which is $d!S(q,d)$ with $S(q,d)$ a Stirling number of
                the second kind.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 24 at 6:41









                Lord Shark the UnknownLord Shark the Unknown

                106k1161133




                106k1161133






























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