Partition of a set into some subsets which are not disjoint necessary [closed]

Let B={1,2,3,...,10}. If we select the subsets A1,...,A6, where B=\Union A1...\Union A6, which A1,.., A6 are not disjoint sets necessarily. Then in how many ways we can select these 6 subsets?

• This is called a cover, not a partition. Is $A_1=B$ and all other $A_i = \emptyset$ considered to be a different "way" than $A_2=B$ and all other $A_i = \emptyset$? May 25 '21 at 21:03
• You've presented a problem without context, making it difficult for Readers to appreciate your interest and level of prior work on the problem. The terse problem statement also fails to clarify some aspects, such as whether the subsets $A_i$ are necessarily distinct and whether they can be empty. Please edit the body of your Question. May 25 '21 at 21:03

Let M be the MULTISET union of A1,...,A6

Suppose $$M$$ has $$i_1$$ 1's, $$i_2$$ 2's,..., $$i_{10}$$ 10's.

Then there are $$\binom{6}{i_1} * \binom{6}{i_2} * ... * \binom{6}{i_{10}}$$ different possible configurations of (A1,...,A6) that result in such an M.

As a result, we see that the number of ways of selecting these 6 subsets is

$$\sum_{1<=i1,...,i10<=6}{[\binom{6}{i_1} * \binom{6}{i_2} * ... * \binom{6}{i_{10}}]}$$

$$=(\binom{6}{1} + \binom{6}{2} + ... + \binom{6}{10})^{10}$$

$$=(2^6-1)^{10}$$

• A more direct argument is that each element of $B$ appears in some nonempty subset of $\{1,\dots,6\}$, and these choices are independent. May 25 '21 at 21:08
• yeah, writing out that last expression made me realize that May 25 '21 at 21:14