# Is this true for every partitioning?

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.

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