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[Problem Solving]

试题详情

题目:

Of the three-digit positive integers that have no digits equal
to zero, how many have two digits that are equal to each
other and the remaining digit different from the other two?

选项:

A、24
B、36
C、72
D、144
E、216

答案:

E

提问:

请问老师这题是不是要分三种情况考虑? AAB ABA BAA这样? 请问具体怎么做?

解答:

点赞0
阅读3474
解答: sysadmin

提问:

请问May老师,这道题我选了C,是用9×8×1=72算的。看到网上论坛说,还要再乘以一个3,即9×8×1×3=216,因为这三个数字是乱序排列的,我不是太懂。我觉得,用9×8×1已经包含了所有顺序的情况了,而且乘以3是什么意思呢?请老师解答。另外,有没有其他更好理解的方法呢?

解答:

点赞3
阅读3644
解答: 徐小梅老师

提问:

请问老师这题是不是要分三种情况考虑? AAB ABA BAA这样? 请问具体怎么做?

解答:

点赞0
阅读3475
解答: sysadmin老师

提问:

看了其他同学解释,没看懂

解答:

点赞0
阅读3441
解答: sysadmin老师

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