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i think the answer is 60.. phew!! tough one.. i dont know whether my approach is correct.. C&C welcome..
starting from ground up .. consider the simplest scenario --------- 2 pairs of siblings - (ab) (cd)
possible team combinations: a c + b d & a d + b c
the way I look at it is .. a[c/ d] = two ways makes sense.. if it doesnt .. pls stop reading here.. rest will all be gibberish.. --------
3 pairs of siblings - (ab) (cd) (ef ) possible combinations
a [c/ d/ e/ f] = 4 ways
remaining players are always similar to (b) (d) (ef) which can be arranged in 2 ways in turn
=> two players without their siblings and one sibling pair
so total = 8 ways
4 pairs of siblings (ab) (cd) (ef) (gh)
a [c/ d/ e/ f/ g/ h] = 6 ways and remainng will be similar to
(b) (d) (ef) (gh)
I can arrange them by picking the first two (bd) and rearrange the remaining 4 into two teams = 2 ways or I can pick one single sibling and other from sibling pair like b [e/ f/ g/ h] and with d in two ways = 8 ways