Let A be a set of n elements.The number of ways we can choose and ordered pair (B,C) where B,C are disjoint subsets of A equals:
A) $n^2$ B)$n^3$ C)$2^n$ D) $3^n$
Here my answer is coming to D) $3^n$ because every single element has 3 possibilities of going either to B or C or none of B or C
But Lets check my answer with A containing 3 elements A={1,2,3} The disjoint subsets can be a) B={1}, C={2} and vice versa
b) B={1}, C={3} and vice versa
c) B={2}, C={3} and vice versa
So there are 6 possibilities of ordered pair (B,C). Now coming to 2 & 1 possibilities
d) B={1,2} C={3} and vice versa
e) B={1,3} C={2} and vice versa
f) B={2,3} C={1} and vice versa
So there are 6 possibilities of ordered pair (B,C). Finally 3 & 0 possibilities
g) B={1,2,3} C= null set and vice versa
So there are two possibilities.
Total there are 6+6+2+1(B=null,C=null) = 15 possibilities
But from my answer it should come to $3^3$= 27 possibilities
Where am I going wrong? Please help
$\{1,2,3\}$ has $8$ subsets:
$$ \{\},\\ \{1\},\\ \{1,2\},\\ \{2\},\\ \{1,3\},\\ \{2,3\},\\ \{1,2,3\},\\ \{3\}, $$ of sizes $0,1,2,1,2,2,3,1$.
Then number of disjoint subsets are respectively
$$8,4,2,4,2,2,1,4$$ (i.e. $2^{3-\#S_i}$) the sum of which $27$.
More generally, the number of subsets of size $k$ are $\displaystyle\binom nk$, leading to the total number of subset combinations
$$\sum_{k=0}^n\binom nk 2^k=(2+1)^n.$$