Counting Pairs with Common Elements in a Set

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SUMMARY

The problem involves counting pairs of subsets A and B from the set X = {1,2,3,4,...,10} such that A ≠ B and A ∩ B = {5,7,8}. The correct approach to solve this is to recognize that each of the remaining 7 elements (1,2,3,4,6,9,10) can be placed in three categories: in A only, in B only, or in neither. This results in 3^7 total combinations for the remaining elements, leading to the conclusion that the total number of valid pairs is 3^7 - 1.

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physics kiddy
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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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