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Perms and Combs problem 
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#1
Jan213, 11:18 AM

P: 135

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 2^71 such numbers A and B. However the answer is 3^7  1. I don't know how .... 


#2
Jan213, 03:18 PM

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P: 9,852

There would be 2^{7} 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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