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Dice with different orders are considered as the same variant. (die1, die2) = (die2, die1).
n=10 m=3 ans=294197
Hint: Generating all combinations of 3 dices works for this problem in Python. At first I thought there must be some other approach to pass all test cases, but it turns out if you find the right way to implement your combinations loop which avoids iterating through duplicate choices of dices, it would be fine.
When there are more than 1 cube into consideration and say for the first cube the number of possibilities are 70 and for the second cube the number of possibilities are 126.
Shouldn't the number of distinct arrangements be 70*126. I am bit stuck up at this point. Can someone give me a sample testcase for more than one cube in consideration
A careful reading of the example shows that the dice are not in fixed position, e.g. one die has 0 so must be first for 01, but (only) the other die has 1 (so the other die is first for 16).
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