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The digits following after the very first digit are 0, 1, 2 and 3 and all their permutations.
Notice that the sum 0 + 1 + 2 + 3 = 6 is less than 10,
while 1 + 2 + 3 + 4 = 10, so the digits 1, 2, 3 and 4, starting from 1, do not satisfy the condition.
Therefore, you have 4! = 24 options by making permutations of the digits 0, 1, 2 and 3 in the 2-nd, 3-rd, 4-th and 5-th positions.
In the very first position you have the digit 6 in all cases.
Solved.