got it! @sagar there shudn be 2's anywhere in the expression. The different combinations formed on the left hand side by choosing 6 out of 12 will ensure different combinations of other 6 people on the right. So 2*12C6 is not required. Example: P1,P3,P5,P7,P9, P11 on left will leave P2, P4, P6, P8, P10 and P12 on right. If 2*12C6 is done, this combination is counted again.
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