I think it would be : (9/10)*10C5*(5C2*(7!/(2!*2!))+5C1*(7!/3!))

9/10 takes care of leading 0 ( since all the digits have equal chance of being leading digit

10C5 selects first 5 different digits

5C2 selects 2 different digits from already used 5 digits to make the number look like (aabbcde)

7!/2!2! is the number of permutations for such numbers (aabbcde)

5C1 selects 1 digit from already used 5 digits to make the number look like (aaabcde)

7!/3! is the number of permutations for such numbers(aaabcde)

hence the final answer is (9/10)*10C5*(5C2*(7!/(2!*2!))+5C1*(7!/3!))

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