Project Euler #90: Cube digit pairs

  • + 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