How many ways are there to choose k numbers from first n natural numbers such that any two numbers are at...












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How many ways are there to choose $k$ distinct numbers from $n$ consecutive natural numbers such that any two numbers are at most differ by $d$?
The solution of the similar problem is done when difference is at least $d$.But what about when the difference is at most $d$?










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    $begingroup$


    How many ways are there to choose $k$ distinct numbers from $n$ consecutive natural numbers such that any two numbers are at most differ by $d$?
    The solution of the similar problem is done when difference is at least $d$.But what about when the difference is at most $d$?










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      $begingroup$


      How many ways are there to choose $k$ distinct numbers from $n$ consecutive natural numbers such that any two numbers are at most differ by $d$?
      The solution of the similar problem is done when difference is at least $d$.But what about when the difference is at most $d$?










      share|cite|improve this question









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      How many ways are there to choose $k$ distinct numbers from $n$ consecutive natural numbers such that any two numbers are at most differ by $d$?
      The solution of the similar problem is done when difference is at least $d$.But what about when the difference is at most $d$?







      combinatorics






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      asked Jan 7 at 12:35









      Supriyo HalderSupriyo Halder

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      635113






















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          $begingroup$

          Assume WLOG that the numbers are $1, 2, ..., n$ and assume $k-1leq d<n$.



          Let's focus on numbers $1, 2, ..., d+1$. Obviously, the difference between any two numbers is at most $d$. We can choose $k$ numbers in $binom{d+1}{k}$ ways.



          Now focus on numbers $2, 3, ..., d+2$. Since we have already considered all $k-subsets$ on the first $d$ numbers, we have to include $d+2$ in our new choices, hence, leading to $binom{d}{k-1}$ ways.



          Focus on $3, 4, ..., d+3$. Same logic as above.



          Repeat this $n-d$ times and get a total of



          $$underbrace{binom{d+1}{k} + binom{d}{k-1} + ... + binom{d}{k-1}}_text{n-d terms}=binom{d+1}{k}+(n-d-1)binom{d}{k-1}$$.






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            1 Answer
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            $begingroup$

            Assume WLOG that the numbers are $1, 2, ..., n$ and assume $k-1leq d<n$.



            Let's focus on numbers $1, 2, ..., d+1$. Obviously, the difference between any two numbers is at most $d$. We can choose $k$ numbers in $binom{d+1}{k}$ ways.



            Now focus on numbers $2, 3, ..., d+2$. Since we have already considered all $k-subsets$ on the first $d$ numbers, we have to include $d+2$ in our new choices, hence, leading to $binom{d}{k-1}$ ways.



            Focus on $3, 4, ..., d+3$. Same logic as above.



            Repeat this $n-d$ times and get a total of



            $$underbrace{binom{d+1}{k} + binom{d}{k-1} + ... + binom{d}{k-1}}_text{n-d terms}=binom{d+1}{k}+(n-d-1)binom{d}{k-1}$$.






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              2












              $begingroup$

              Assume WLOG that the numbers are $1, 2, ..., n$ and assume $k-1leq d<n$.



              Let's focus on numbers $1, 2, ..., d+1$. Obviously, the difference between any two numbers is at most $d$. We can choose $k$ numbers in $binom{d+1}{k}$ ways.



              Now focus on numbers $2, 3, ..., d+2$. Since we have already considered all $k-subsets$ on the first $d$ numbers, we have to include $d+2$ in our new choices, hence, leading to $binom{d}{k-1}$ ways.



              Focus on $3, 4, ..., d+3$. Same logic as above.



              Repeat this $n-d$ times and get a total of



              $$underbrace{binom{d+1}{k} + binom{d}{k-1} + ... + binom{d}{k-1}}_text{n-d terms}=binom{d+1}{k}+(n-d-1)binom{d}{k-1}$$.






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              $endgroup$
















                2












                2








                2





                $begingroup$

                Assume WLOG that the numbers are $1, 2, ..., n$ and assume $k-1leq d<n$.



                Let's focus on numbers $1, 2, ..., d+1$. Obviously, the difference between any two numbers is at most $d$. We can choose $k$ numbers in $binom{d+1}{k}$ ways.



                Now focus on numbers $2, 3, ..., d+2$. Since we have already considered all $k-subsets$ on the first $d$ numbers, we have to include $d+2$ in our new choices, hence, leading to $binom{d}{k-1}$ ways.



                Focus on $3, 4, ..., d+3$. Same logic as above.



                Repeat this $n-d$ times and get a total of



                $$underbrace{binom{d+1}{k} + binom{d}{k-1} + ... + binom{d}{k-1}}_text{n-d terms}=binom{d+1}{k}+(n-d-1)binom{d}{k-1}$$.






                share|cite|improve this answer









                $endgroup$



                Assume WLOG that the numbers are $1, 2, ..., n$ and assume $k-1leq d<n$.



                Let's focus on numbers $1, 2, ..., d+1$. Obviously, the difference between any two numbers is at most $d$. We can choose $k$ numbers in $binom{d+1}{k}$ ways.



                Now focus on numbers $2, 3, ..., d+2$. Since we have already considered all $k-subsets$ on the first $d$ numbers, we have to include $d+2$ in our new choices, hence, leading to $binom{d}{k-1}$ ways.



                Focus on $3, 4, ..., d+3$. Same logic as above.



                Repeat this $n-d$ times and get a total of



                $$underbrace{binom{d+1}{k} + binom{d}{k-1} + ... + binom{d}{k-1}}_text{n-d terms}=binom{d+1}{k}+(n-d-1)binom{d}{k-1}$$.







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                share|cite|improve this answer










                answered Jan 7 at 13:10









                EuxhenHEuxhenH

                472210




                472210






























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