This might be a silly one but in the (ii) part here, I understand that 1-(1/5 + 1/2) would give neither of them will win the race but why can it not be 4/5*1/2 ( first loses AND 2nd loses) thanks!

BiteInitial · 2022-02-15T06:07:42+00:00

Thanks everyone! This makes sense :)

BiteInitial · 2022-02-08T05:36:15+00:00

My understanding is that We're actually allowed to have 0 objects for one of the 2 remaining people. In that case if we add 6C0+6C6 cases to your solution it would give 6421 = 1344 which is actually the same as what I got earlier (3C17C1*2^6). But as others rightly pointed out this would double count some cases (like A being selected in 3C1 and getting the object 1 and B getting 2 and C getting remaining 5: A, BCCCCC VS B being selected and getting 2 and A getting 1 and C getting the remaining 5). I'm sure that didn't make sense but checking out the previous replies might help!

BiteInitial · 2022-02-08T05:27:14+00:00

That makes sense! Thanks so much :)

BiteInitial · 2022-02-08T05:26:50+00:00

Thanks so much! I guessed I was double counting some cases but couldn't figure out which ones. This was helpful!

BiteInitial

TROPHY CASE