The "Brilliant" Set

Let $P_{n,m}$ denote the number of ways of partitioning $n$ numbers in $m$ non-empty and disjoint subsets. Clearly, we have $P_{n,1}=1, \hspace{10pt} \forall n \geq 1,\hspace{56pt}(1)$ Now consider adding a new element to an existing partition with $n-1$ elements and $m$ subsets. This new element can go into any one of the previous $m$ subsets (thus preserving the number of subsets ) or form a subset of its own (and hence increasing the number of subsets from $m$ to $m+1$ ). Thus, we have the following recursions : $P_{n.m}=mP_{n-1,m}+P_{n-1,m-1}, \hspace{10pt} \text{if} \hspace{5pt}m\leq n$ and, $P_{n,m}=0\hspace{10pt}\text{if}\hspace{10pt} m >n \hspace{20pt}(2)$ The recursions (1) and (2) can be solved easily. In our problem, we have $n=7$ and we require $\sum_{m=1}^{7}P_{7,m}$ , which is computed to be $877$ .

The number of ways of partitioning a set of $n$ distinct elements is the Bell number $B_{n}.$ From the sequence given here we see that the answer to this question is $B_{7} = \boxed{877}.$

Solve using 'Atkins array'

That is

$a$

$a,2a$

$2a,3a.5a$

$5a,7a,10a,15a$

$............$

$.............................$

So, in all cases we have $a=1$ and the $\text{First element of(which is same as the last element of}$ $(n-1)^{th}$ ) row $n^{th}$ $\text{row represents} A_{n-1}$ where $A_{n-1}$ is the number of partitions of set (containg $n$ distinct elements) into disjoint subsets whose union is A

So we have the atkins array as

$1 = A_0$

$1,2=A_1$

$2,3,5=A_2$

$5,7,10,15=A_3$

$15,20,27,37,52=A_4$

$52,67,87,114,151,203=A_5$

$203,255,322,409,523,674,\boxed{877}=A_6$

SO $\boxed {877}$ is the answer, As the first element of $8^{th}$ is same as last element of $7^{th}$ row.

There is no direct solution yet. We can simply calculate the all possible configurations of splitting, see the picture:

The rows are reading as follows: the number of elements at every subset (italic). The bold number is the number of possible configuration.

Just for example the row H $(3,3,1)$ is calculated as follow: $\frac{\binom{7}{3}*\binom{7-3}{3}}{2!}=70$

We need to divide by the possible rearrange of subsets of $3$ .

Repeating for every configuration we have the total result: $877$