In case you don't want to leave it as an application of some 'Bell Number', here's a more intuitive solution.
We can partition 5 symbols, ie {A, C}, B, D, E, F in 5 ways that is,
All of them in 1 partition
All of them divided in 2 partitions
Similarly 3, 4, 5 partitions
1 and 5 partitions:
Either have all of them in one partition or each separately. Hence total = 2
2 partitions:
Can be divided as 2 elements in one partition and remaining 3 in the other OR 1 element in one partition and remaining 4 in the other. Ways of doings it = $\frac{5!}{3!2!} + \frac{5!}{1!4!} = 10+ 5= 15$
3 partitions:
Partitions can be formed in 2 ways. Either {1, 1, 3} or {1, 2, 2} . So ways of doing it = $\frac{5!}{3!2!} + \frac{5!}{2!2!2!} = 10+ 15= 25$
4 partitions:
Only one possible way to divide which is {1, 1, 1, 2}. Hence $\frac{5!}{1!1!1!2!3!} = 10$
Adding all of them =>
$2+15+25+10=52$