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Perms and Combs problem

  1. Jan 2, 2013 #1
    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 ....
     
  2. jcsd
  3. Jan 2, 2013 #2

    haruspex

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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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