I just would like to thank you all for coming out with a solution for the puzzle I presented , i think you are all top mathmaticians and programmers. Now i am using it to take the pixels that are least common and and least ifferent form 2 pictures. I think all you know how important it is to take out the sequences a it applies to many areas of imaging. Unfortunately i do not know how to apply it tomodifying image pixels so that i get an image from imagelevels or image data.

This can be made into a two-liner, generical counting for subsets

In[98]:= (* for the WC *)
a0 = {3, 4, 5, 7, 10, 15, 23, 27};
a1 = {4, 6, 7, 8, 9, 10, 12, 15, 25};
a2 = {2, 6, 7, 10, 11, 12, 13, 14, 15, 25};

In[101]:= Clear[cp]
          cp = Complement[#, Intersection[a0, a1, a2]] & /@ {a0, a1, a2}
Out[102]= {{3, 4, 5, 23, 27}, {4, 6, 8, 9, 12, 25}, {2, 6, 11, 12, 13, 14, 25}}

In[103]:= First /@ (Complement[#, 
     Flatten[Intersection @@@ (Part[cp, #] & /@ 
         Subsets[Range[Length[cp]], {2, Length[cp] - 1}])]] & /@ cp)
Out[103]= {3, 8, 2}

but from the (insinuated) point of view of the problem owner it might consist a problem, that not all subset intersections are non-empty

In[105]:= Intersection @@@ (Part[cp, #] & /@ Subsets[Range[Length[cp]], {2, Length[cp] - 1}])
Out[105]= {{4}, {}, {6, 12, 25}}

which has been flattened away,

Moreover I think it suffices to look at all pairs of subsets of cp

This means that for let's say 4 non-empty sets $A,B,C,D$ the sets

$A_2 = A\cap B \vee A\cap C \vee A\cap D \vee B\cap C \vee B\cap D \vee C\cap D$

and

$A_{23} = A\cap B \vee A\cap C \vee A\cap D \vee B\cap C \vee B\cap D \vee C\cap D \vee A\cap B\cap C \vee A\cap B\cap D \vee A\cap C\cap D \vee B\cap C\cap D$

are identical $A_2 \equiv A_{23}$, which is clear because of the $\vee$ (logical OR) and the fact, that intersections of three sets are not more permissive as intersections of two sets. Actually I was driven by your other remark

Now again find elements in cp common to all members in cp comparing all subsets

to deliver a generic code sample which could deliver all subsets. It was restricted to the subsets with $2,3,...,n-1$ members and should be restricted even more in the given context, as you say.

But what does the OR between two sets (here intersections) mean?

What do you think does it mean? That $A_2$ consists of elements from $A \cap B$ (inclusively) or from $A \cap C$ ... but this must be correctly written as $A_2 = (A \cap B) \cup (A \cap C) \cup \dots$ which means with respect to $A_2$ and $A_{23}$ that $A_2 = A_{23}$ because $A\cap B \cap C \subseteq (A \cap B) \cup (A \cap C)$ and $A\cap B \cap C \subseteq (B \cap C) \cup (B \cap A)$ and $A\cap B \cap C \subseteq (C \cap A) \cup (C \cap B)$ and so on and so on.

Intersection has to be applied usually to a sequence of lists, but as mostly often

Intersection works with any head, not just List:

So, if something isn't an atom

In[8]:= AtomQ[a[1]]
Out[8]= False

it has a head and this head remains unchanged, the content is intersected - as it is done with List[]:

In[8]:= AtomQ[a[1]]
Out[8]= False

In[9]:= a[1] \[Intersection] a[2]
Out[9]= a[]

In[10]:= {1} \[Intersection] {2}
Out[10]= {}

If the heads do not fit, you get it to know

In[11]:= a[1, 2, 3, 4] \[Intersection] a[2, 3, 4, 5] \[Intersection] b[1, 2, a[1, 2, 3, 4]]
During evaluation of In[11]:= Intersection::heads: Heads b and a at positions 3 and 1 are expected to be the same. >>
Out[11]= a[1, 2, 3, 4] \[Intersection] a[2, 3, 4, 5] \[Intersection] b[1, 2, a[1, 2, 3, 4]]

it's once again a reflection of the deep insight that heads are not as much different from each other as they appear in education for childs and students:

In[13]:= a[1, 2, 3, 4] \[Intersection] a[2, 3, 4, 5] \[Intersection]  a[1, 2, b[1, 2, 3, 4]]
Out[13]= a[2]

If your a2 is corrected (a 23 between 12 and 13 has been canceled) then

In[11]:= a0 = {3, 4, 5, 7, 10, 15, 23, 27};
         a1 = {4, 6, 7, 8, 9, 10, 12, 15, 25};
         a2 = {2, 6, 7, 10, 11, 12, 13, 14, 15, 25};

In[8]:= Plus @@ Most[a0]
Out[8]= 67

In[9]:= Plus @@ Most[a1]
Out[9]= 71

In[14]:= Plus @@ Most[a2]
Out[14]= 90

In[16]:= 32/13 // N
Out[16]= 2.46154

A has another sum as hand-written and 32/13 is of course not bigger than 24 ... then: in the last sequence C: the difference between 2 and 6 is 4, so 2 and 6 are more different than 7 and 10, why 7 and 10 have been selected?