# Project Euler #90: Cube digit pairs

# Project Euler #90: Cube digit pairs

+ 0 comments I would make something more explicit.

one way this can be achieved is by placing 056789 on one cube and 123489 on the other cube

One could think that it counts as 2 solutions:

- 056789 on the first one, 123489 on the second
- 123489 on the first one, 056789 on the second

This is not the case. It counts as 1 solution.

So for 9 2 the answer is 1217.

For 1 2 the answer is 13461. It can be also computed manually:

0 1 3136 0 01 3920 01 01 2485 01 1 3920

+ 0 comments If the question had used the term

**dice**instead of**cube**it would have been fun to add the input`S`

for the number of sides on the dices.A single cube only has 210 arrangements.

+ 0 comments 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.

+ 0 comments Dice with different orders are considered as the same variant. (die1, die2) = (die2, die1).

n=10 m=3 ans=294197

+ 2 comments Hey

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

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