Is this true for every partitioning?

Question

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.

Answer 1

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 $$\sum_{t=1}^T \dbinom{t}{2} n_t=6<7=\sum_{t=1}^T \dbinom{t}{2} n'_t.$$