Counting Pairs with Common Elements in a Set

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The problem involves finding the number of pairs {A, B} where A and B are subsets of X = {1,2,3,4,...,10}, with the condition that A ≠ B and A ∩ B = {5,7,8}. After identifying the remaining elements {1,2,3,4,6,9,10}, which total 7, the initial approach suggests using permutations to distribute these elements into two sets. However, the correct calculation involves recognizing that each of the 7 elements can be placed in A, B, or neither, leading to 3^7 total combinations. The final result is derived by subtracting the case where A and B are identical, yielding 3^7 - 1 valid pairs.
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Here's the problem :

Let X = {1,2,3,4 ... 10}. Find the number of pairs {A,B} such that A \subseteq X and B \subseteq X, A \neq
B and A \cap B = {5,7,8}.

My attempt:

Once we know that the remaining numbers are 1,2,3,4,6,9,10 ... a total of 7 numbers, we can use permutation to know that seven elements can be distributed to 2 sets in 2^7 ways ...

Excluding A and B having the common elements {5,7,8}, we have a total of 2^7-1 such numbers A and B.

However the answer is 3^7 - 1. I don't know how ...
 
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There would be 27 ways of placing each of the 7 remaining digits in either A or B. But they need not be in either.
 
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