Perms and Combs problem

physics kiddy

Here's the problem :

Let X = {1,2,3,4 ...... 10}. Find the number of pairs {A,B} such that A [itex]\subseteq[/itex] X and B [itex]\subseteq[/itex] X, A [itex]\neq[/itex]
B and A [itex]\cap[/itex] 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 ....

haruspex

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.