A probability problem by EKENE FRANKLIN

2013 2013 is the first year in the Middle Ages that consists of four consecutive digits. The number of such years that will occur after 2013 2013 and before year 10000 10000 is x x . Find x + 6 x + 6 .

256 155 7997 1997

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1 solution

Parth Sankhe
Oct 12, 2018

For 0, 1, 2, 3, the years after 2013 would be 2103 , 2130 , 2031 , 2301 , 2310 2103,2130,2031,2301,2310 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 ! 4! permutations.

Thus the total number of required years is 5 + 6 + 6 × 4 ! = 155 5+6+6×4!=155

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