Question 1117956
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The members of the committee go from different twin sets.


So, there are  {{{C[10]^3}}} choices between the sets of twins, and then 2*2*2 choices inside each twin set.


The total number of ways is  {{{C[10]^3*8}}} = {{{((10*9*9)/(1*2*3))*8}}} = 1080.
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