From Schaum's Discrete Math:
Assuming a cell can be empty find the number of ways a set of 3 elements can be partitioned into:
a) 3 ordered cells b) 3 unordered cells
So let's say I've got the set ${a,b,c}$. So I have 1 way to choose a set of 3 and then 3 ways to arrange that set in 3 distinct cells. I then have $\binom{3}{2}=3$ ways to choose a set of 2 which then gives me a set of 2, a set of 1, and the empty set which can be arranged in $3!$ ways. I then have $3!$ ways to arrange the elements as single element sets. Thus I end up with $3+3*6+6=27$
However they just answered a) $3^3=27$ and b) 5 without any explanation.
How did they come up with $3^3$? How do you explain that?