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#31
Arc, you are my idol! (Sorry Julian, it's Arc's turn now.) You seem to be mathematically very talented :).
Arc, you are my idol! (Sorry Julian, it's Arc's turn now.) You seem to be mathematically very talented :).
I am not very talented, Kari :) As I can say, I am just above average.
Very talented Indians live in USA
Well I don't know. This thread is marked as solved but three different answers have been given: 4, 8 and "2!3".
2!3 is wrong, because 2!=2 and 3=3, so 2!3=2 x 3 which is 6.
The answer depends on how you specify the question. If the sequence of numbers is important, then indeed there are 8 possible outcomes. If the sequence is not important (and throwing 3 disks randomly up in the air to see how they land rather implies that it isn't), then there are only 4 possible combinations of numbers showing.
So the answer is probably 4, could arguably be 8, but absolutely definitely is NOT 2!3....
The answer to this enigma is, I believe, just 4. Suppose the objects in question were just simple coins, each comprising a Head and a Tail. We therefore have the following combinations: HHH, HHT, HTT, and TTT (note that HTH and THH are identical combination-wise to HHT and, likewise, so are THT and TTH to HTT).