is the first year in the Middle Ages that consists of four consecutive digits. The number of such years that will occur after and before year is . Find .
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For 0, 1, 2, 3, the years after 2013 would be 2 1 0 3 , 2 1 3 0 , 2 0 3 1 , 2 3 0 1 , 2 3 1 0 and there would be 6 for the years starting with 3.
Now we have 6 ways to choose 4 consecutive integers from 1 to 9, and each will have 4 ! permutations.
Thus the total number of required years is 5 + 6 + 6 × 4 ! = 1 5 5