Of the three-digit integers greater than 700 ,how many have two digits that are equal to each other and the remaining digit different from the other two?a/ 90b/ 82c/ 80d/ 45e/ 36

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Of the three-digit integers greater than 700 ,how many have two digits that are equal to each other and the remaining digit different from the other two?a/ 90b/ 82c/ 80d/ 45e/ 36
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Of the three-digit integers greater than 700 ,how many have two digits that are equal to each other and the remaining digit different from the other two?a/ 90b/ 82c/ 80d/ 45e/ 36
Of the three-digit integers greater than 700 ,how many have two digits that are equal to each other and the remaining digit different from the other two?
a/ 90
b/ 82
c/ 80
d/ 45
e/ 36

Of the three-digit integers greater than 700 ,how many have two digits that are equal to each other and the remaining digit different from the other two?a/ 90b/ 82c/ 80d/ 45e/ 36
这一题先翻译一下：
在大于700的三位整数中,有多少个整数其中有两个位数上的数字相等而剩余的数字不和这个数字相等,例如：833,838,883……注意,由于有一个数不能相等,所以888不是.
既然理解了题目的意思,那么我们可以将相同的归为3类：百位和十位相等,十位和个位相等,百位和个位相等.
当百位和十位相等时,题目限制了百位至少为7,至多为9,所以一共三种可能性：7,8,9.而中间0~9中每一种可能性都会导致一个数是不允许的,也就是777,888,999.所以一共有3×10-3=27种；
当百位和十位相等时,思考方式同上,也是27种（不懂再追问）
当个位和十位相等时,考虑到百位的存在,当百位是7时不能取7,百位是8时不能取8,百位上是9时不能取9,所以也是3*10-3=27种,总共是27×3=81种,那么,没一个选项是81,怎么回事呢?
原来在第三种的时候,将700这种情况也算进去了,所以要再减1,等于80,至此,回答完毕.

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