Is this true for every partitioning?












3












$begingroup$


I have two categories (category1 and category2 ) and The size of both categories is equal to each other. if we partition each categories arbibtrary .Is this proposition proven? or rejected?



$n_T gt n_T' $ $implies$ $sum_{t=1}^T dbinom{t}{2} n_t gt sum_{t=1}^T dbinom{t}{2} n'_t$



consider the number of the partitions with size $t$ as $n_t$.and Suppose that the size of the largest partition is equal to $T$.



for example if size of each categories be equal to 33. we can partition one of them into one partitions with size 5 and one partitions with size 4 and two partitions with size 3 and seven partitions with size 2. (In other words , $n_5=1 , n_4 = 2 , n_3=2 , n_2 = 7$ because $5+2*4+2*3+7*2=33$) , And the size of the largest partition is 5.



now for category1 and category 2 suppose that



$T$ : the size of the largest partition in each category .(The value of $T$ is the same in both categories)



$n_t $: number of the partitions with size $t$ in category1.( $t=1,…,T$)



$n'_t$: number of the partitions with size $t$ in category2.( $t=1,…,T$)



Is this true for every partitioning?



$n_T' gt n_T' $ $implies$ $sum_{t=1}^T dbinom{t}{2} n_t gt sum_{t=1}^T dbinom{t}{2} n'_t$



I consider $dbinom{1}{2} = 0 $



until now I could not find a counterexample.



Thanks a lot for reading.






combinatorics combinations integer-partitions set-partition clustering







share|cite|improve this question









$endgroup$












  • $begingroup$
    Hi @Alex Ravsky Thanks a lot !
    $endgroup$
    – Richard
    Dec 1 '18 at 17:14
















3












$begingroup$


I have two categories (category1 and category2 ) and The size of both categories is equal to each other. if we partition each categories arbibtrary .Is this proposition proven? or rejected?



$n_T gt n_T' $ $implies$ $sum_{t=1}^T dbinom{t}{2} n_t gt sum_{t=1}^T dbinom{t}{2} n'_t$



consider the number of the partitions with size $t$ as $n_t$.and Suppose that the size of the largest partition is equal to $T$.



for example if size of each categories be equal to 33. we can partition one of them into one partitions with size 5 and one partitions with size 4 and two partitions with size 3 and seven partitions with size 2. (In other words , $n_5=1 , n_4 = 2 , n_3=2 , n_2 = 7$ because $5+2*4+2*3+7*2=33$) , And the size of the largest partition is 5.



now for category1 and category 2 suppose that



$T$ : the size of the largest partition in each category .(The value of $T$ is the same in both categories)



$n_t $: number of the partitions with size $t$ in category1.( $t=1,…,T$)



$n'_t$: number of the partitions with size $t$ in category2.( $t=1,…,T$)



Is this true for every partitioning?



$n_T' gt n_T' $ $implies$ $sum_{t=1}^T dbinom{t}{2} n_t gt sum_{t=1}^T dbinom{t}{2} n'_t$



I consider $dbinom{1}{2} = 0 $



until now I could not find a counterexample.



Thanks a lot for reading.






combinatorics combinations integer-partitions set-partition clustering







share|cite|improve this question









$endgroup$












  • $begingroup$
    Hi @Alex Ravsky Thanks a lot !
    $endgroup$
    – Richard
    Dec 1 '18 at 17:14














3












3








3


2



$begingroup$


I have two categories (category1 and category2 ) and The size of both categories is equal to each other. if we partition each categories arbibtrary .Is this proposition proven? or rejected?



$n_T gt n_T' $ $implies$ $sum_{t=1}^T dbinom{t}{2} n_t gt sum_{t=1}^T dbinom{t}{2} n'_t$



consider the number of the partitions with size $t$ as $n_t$.and Suppose that the size of the largest partition is equal to $T$.



for example if size of each categories be equal to 33. we can partition one of them into one partitions with size 5 and one partitions with size 4 and two partitions with size 3 and seven partitions with size 2. (In other words , $n_5=1 , n_4 = 2 , n_3=2 , n_2 = 7$ because $5+2*4+2*3+7*2=33$) , And the size of the largest partition is 5.



now for category1 and category 2 suppose that



$T$ : the size of the largest partition in each category .(The value of $T$ is the same in both categories)



$n_t $: number of the partitions with size $t$ in category1.( $t=1,…,T$)



$n'_t$: number of the partitions with size $t$ in category2.( $t=1,…,T$)



Is this true for every partitioning?



$n_T' gt n_T' $ $implies$ $sum_{t=1}^T dbinom{t}{2} n_t gt sum_{t=1}^T dbinom{t}{2} n'_t$



I consider $dbinom{1}{2} = 0 $



until now I could not find a counterexample.



Thanks a lot for reading.






combinatorics combinations integer-partitions set-partition clustering







share|cite|improve this question









$endgroup$




I have two categories (category1 and category2 ) and The size of both categories is equal to each other. if we partition each categories arbibtrary .Is this proposition proven? or rejected?



$n_T gt n_T' $ $implies$ $sum_{t=1}^T dbinom{t}{2} n_t gt sum_{t=1}^T dbinom{t}{2} n'_t$



consider the number of the partitions with size $t$ as $n_t$.and Suppose that the size of the largest partition is equal to $T$.



for example if size of each categories be equal to 33. we can partition one of them into one partitions with size 5 and one partitions with size 4 and two partitions with size 3 and seven partitions with size 2. (In other words , $n_5=1 , n_4 = 2 , n_3=2 , n_2 = 7$ because $5+2*4+2*3+7*2=33$) , And the size of the largest partition is 5.



now for category1 and category 2 suppose that



$T$ : the size of the largest partition in each category .(The value of $T$ is the same in both categories)



$n_t $: number of the partitions with size $t$ in category1.( $t=1,…,T$)



$n'_t$: number of the partitions with size $t$ in category2.( $t=1,…,T$)



Is this true for every partitioning?



$n_T' gt n_T' $ $implies$ $sum_{t=1}^T dbinom{t}{2} n_t gt sum_{t=1}^T dbinom{t}{2} n'_t$



I consider $dbinom{1}{2} = 0 $



until now I could not find a counterexample.



Thanks a lot for reading.






combinatorics combinations integer-partitions set-partition clustering




combinatorics combinations integer-partitions set-partition clustering






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asked Nov 30 '18 at 10:53









RichardRichard

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265












  • $begingroup$
    Hi @Alex Ravsky Thanks a lot !
    $endgroup$
    – Richard
    Dec 1 '18 at 17:14


















  • $begingroup$
    Hi @Alex Ravsky Thanks a lot !
    $endgroup$
    – Richard
    Dec 1 '18 at 17:14
















$begingroup$
Hi @Alex Ravsky Thanks a lot !
$endgroup$
– Richard
Dec 1 '18 at 17:14




$begingroup$
Hi @Alex Ravsky Thanks a lot !
$endgroup$
– Richard
Dec 1 '18 at 17:14










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If I understood your question right, the answer is negative. Consider two partitions of $11$ such that $n_3=2$, $n_2=0$, $n_1=5$, $n’_3=1$, $n_2=4$, and $n’_1=0$. Then $n_{3}>n’_3$, but
$$sum_{t=1}^T dbinom{t}{2} n_t=6<7=sum_{t=1}^T dbinom{t}{2} n'_t.$$






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    $begingroup$

    If I understood your question right, the answer is negative. Consider two partitions of $11$ such that $n_3=2$, $n_2=0$, $n_1=5$, $n’_3=1$, $n_2=4$, and $n’_1=0$. Then $n_{3}>n’_3$, but
    $$sum_{t=1}^T dbinom{t}{2} n_t=6<7=sum_{t=1}^T dbinom{t}{2} n'_t.$$






    share|cite|improve this answer









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      $begingroup$

      If I understood your question right, the answer is negative. Consider two partitions of $11$ such that $n_3=2$, $n_2=0$, $n_1=5$, $n’_3=1$, $n_2=4$, and $n’_1=0$. Then $n_{3}>n’_3$, but
      $$sum_{t=1}^T dbinom{t}{2} n_t=6<7=sum_{t=1}^T dbinom{t}{2} n'_t.$$






      share|cite|improve this answer









      $endgroup$
















        1












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        $begingroup$

        If I understood your question right, the answer is negative. Consider two partitions of $11$ such that $n_3=2$, $n_2=0$, $n_1=5$, $n’_3=1$, $n_2=4$, and $n’_1=0$. Then $n_{3}>n’_3$, but
        $$sum_{t=1}^T dbinom{t}{2} n_t=6<7=sum_{t=1}^T dbinom{t}{2} n'_t.$$






        share|cite|improve this answer









        $endgroup$



        If I understood your question right, the answer is negative. Consider two partitions of $11$ such that $n_3=2$, $n_2=0$, $n_1=5$, $n’_3=1$, $n_2=4$, and $n’_1=0$. Then $n_{3}>n’_3$, but
        $$sum_{t=1}^T dbinom{t}{2} n_t=6<7=sum_{t=1}^T dbinom{t}{2} n'_t.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 1 '18 at 14:25









        Alex RavskyAlex Ravsky

        39.5k32181




        39.5k32181






























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