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- #1

I tried solving this like so,

the first dice has a possible 6 numbers, the second has a possible 5, the third has a possible 4, and the fourth, 3. Then there are 2 remaining dice of which each has to be one of the previous 4 numbers so there are:

6*5*4*3*4*4 = 5760 ways

something tells me I am not thinking correctly or might be missing something because when I did this by iteration and got 9216. (though I might have missed something here too)