if we have a set {a,b,c,d} and two indistinct boxes
To allocate {a,b,c,d}...two-and-two, in those two boxes we have only 3 choices. $\frac{4!}{2!2!}*\frac{1}{2!} = 3$
those are $\left \{ \left \{ a,b \right \} ,\left \{ c,d \right \} \right \} , \ \left \{ \left \{ a,c \right \} ,\left \{ b,d \right \} \right \} , \ \left \{ \left \{ a,d \right \} ,\left \{ b,c \right \} \right \}$
No need to put a in box2 and count again. because boxes are indistinct.
In analogy to these boxes, those groups of a partition are also same,only the elements inside a group are distinct.