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1.There will be 10C2 pairs to play elimination games.
So there will be. 10!/(8!*2!)=10*9/(1*2)=90/2=45 elimination games.
2. 7C3=35 , 7C4=35 ,7C5= 21, 7C6=7 , 7C7=1
Add the answer from 7C3 to 7C7 = 35+35+21+7+1 = 99 polygons
3. The 1st choice is 1 of 52.
Then 1 of 51, etc.
- 52×51×50×49×48 = 311857200
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But, choosing cards A,B,C,D & E is the same as B,A,D,E & C.
= 120 (5×4×3×2×1) ways to get the same 5 cards.
311857200/120 = 2598960 sets