Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

up vote 1 down vote accepted

OK, I get it. It's the number of functions from the set $\{a,b,c\}$ to the set of labelled cells. That's why it's $3^3$. So in the general case it would be $m^n$ to distribute $n$ elements in to $m$ labeled cells.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.