How many ways are there to arrange the integers 0,1,2..,9, so that the first digit is greater than 1 and the last digit is smaller than 7?

First digit can be filled in 8 ways with 2,3,4,5,6,7,8,9. Last digit can be filled in 7 ways with 0,1,2,3,4,5,6

There are 8 remaining places and 10 digits. That can be done in 10^{8 ways.}

Thus 10^{8} * 8* 7 = 56*10^{8} numbers

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