How many ways there are to divide $5$-element set to at mot three sets?

Question

How many ways there are to divide $5$-element set to at most three subsets? If I am right, then I have following subsets:

1 subset: containing five elements (that is 1 possibility)

2 subsets:

• 1 element and 4 elements (that is ${5 \choose 1}\cdot {4 \choose 4}$ possibilities)
• 2 elements and 3 elements (that is ${5 \choose 2}\cdot {3 \choose 3}$ possibilities)

3 subsets:

• 1 el. 1 el. and 3 el. (that is ${5 \choose 1}\cdot {4 \choose 1}\cdot {3 \choose 3}$ possibilities)

• 2 el. 2 el. and 1 el. (that is ${5 \choose 2}\cdot {3 \choose 2}\cdot {1 \choose 1}$ possibilities)

Now I just need to sum that all up. Did I solve it correctly?

Answer 1

There are problems with the three subsets analysis.

We count the number of splittings of type $1$-$1$-$3$. The objects that form the $3$-set can be chosen in $\binom{5}{3}$ ways. Now it's over!

Alternately, the $2$ objects that will be lonely can be chosen in $\binom{5}{2}$ ways. Now it's over.

Similarly, you have double-counted the splittings of type $2$-$2$-$1$.

Remark: Your $\binom{5}{1}\binom{4}{1}$ counts the splittings $\{A\}, \{B\}, \{C,D,E\}$ and $\{B\}, \{A\}, \{C,D,E\}$ as different.

We have left the correction of the $2$-$2$-$1$ count to you. Please leave a message if there ia difficulty.

Answer 2

For two singletons and one subset of $3$ elements there are $\binom{5}{2}=10$ possibilities. So not $20$. You count the possibilities twice.

For one singleton and two subsets of $2$ elements you make a sortlike mistake.