Trenters4325 said:
Let S be a set with six elements. In how many different ways can one select two not necessarily distinct subsets of S so that the union of the two subsets is S? The order of selection does not matter; for example, the pair of subsets {a,c},{b,c,d,e,f}represents the same selection as the pair {b,c,d,e,f},{a,c}.
I think that you mean two not necessarily
disjoint subsets of S.
Let’s agree on some notation. \(\displaystyle S = \{ a,b,c,d,e,f\}\).
If \(\displaystyle T = \{ a,c,d\} \quad \Rightarrow \quad S\backslash T = T^c = \{ b,e,f\}\).
So note that, \(\displaystyle T \cup T^c = S\); and \(\displaystyle U^c \subseteq V\quad \Rightarrow \quad U \cup V = S\).
Now
be very careful, do not over count!
Here is one pair: \(\displaystyle \left\langle {\emptyset ,S} \right\rangle\).
To count the pairs \(\displaystyle \left\langle {\{ a\} ,V} \right\rangle\) V can be any subset of S that contains the complement of {a}.
There are just two possible V’s: {b,c,d,e,f} & S.
Thus for each one element subset, there are two subsets that can be paired with it.
This gives us 12 more. (So far we have 13.)
To count the pairs \(\displaystyle \left\langle {\{ a , b\} ,V} \right\rangle\) V can be any subset of S that contains the complement of {a,b}.
There are just four possible V’s! What are they?
Thus each of the 15 two-element subsets, there are four subsets that can be paired with it. How many more does that give us?
To count the pairs \(\displaystyle \left\langle {\{ a , b , c\} ,V} \right\rangle\) V can be any subset of S that contains the complement of {a,b,c}.
There are just eight possible V’s! WHY?
Now at this point remember: be very careful, do not
over count!
There are 20 three-element subsets. BUT we have already counted some of them because the complement of any three-element subset is a three-element subset.
Hence how many of these parings are there if we avoid the over count.
Now there are three more cases for you to consider.
Four-element subsets paired with 4, 5 & 6 element sets.
Five-element subsets paired with 5 & 6 element sets.
And S itself.
Again in each case,
be very careful, do not over count!
