Question

一个GRE 数学题

Josh works on the second floor of a building. there are 10 doors to the buiding and 8 staircases from the first to the second floor. Josh decided that each day he would enter by one door and leave by a different one, and go up one staircase and down another. how many days could josh do this before he had to repeat path he had previously taken?

Answer 1

5040天
英语没那么好，不介意我用汉字吧，你用英语整理下吧
这是一种数学中比较简单的排列组合，10 个大门，8间楼梯，每天进门出门走不同的门，上下楼经不同的楼梯
首先我们只考虑进出门的方式，进去有10种选择，出来有9种了，所以是9*10=90种。你可以理解为把 10 个各不相同的石头，放进AB两个盒子里，一共有多少种放法，
同样上下楼的的方式，8间楼梯上去8种，那么下来就有7种，7*8=56。 你也可以理解为把1-8八个数字组成一个两位数，一共有多少种方法。
所以最后结果，进出门方式乘以上下楼方式，90*56=5040种

Answer 2

1.we think about the door example:Door1 to 10 he could enter 1 leave 2-10 the same for enter door2 leave 1、3-10 so it has 10*9=90 ways
2.we think about the staircases example: S1-8 he could go up S1 down S2-8 like the 'door' so it has 8*7=56 ways
3.when he use one way to enter/leave the door he has 49 ways to go up / down the staircases so he has 49*81=5040 ways

Answer 3

1个门每次走不同楼道可以走4天，所以10个门可以走40天，则第41天必然要重复前面的路。
所以before he repeat the path应该是40天

Answer 4

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