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A three digit number $n$ is such that the last two digits of it are equal and differ from the first. The number of such $n$ 's is
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The correct answer is:
81
If the last two digits is 0,0 , then in Ist digit any of the numbers except 0 i.e., 9 numbers
If the last two digits is 1,1 , then in Ist digit any of the numbers except 0 and 1 , i.e. 8 numbers
$\therefore$ The total number of numbers
$=9+8 \times 9=81$
If the last two digits is 1,1 , then in Ist digit any of the numbers except 0 and 1 , i.e. 8 numbers
$\therefore$ The total number of numbers
$=9+8 \times 9=81$
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