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Finding 12 as sum of digits between the numbers 1 and 1,000,000 is like finding coefficient of x^12 between 1 and 999,999 (as the digits of course of 1,000,000 don't sum up to 12).

Now all the numbers a1+a2+a3+a4+a5+a6 (number of digits in 999,999) should be between 0 and 9.

So, this is equivalent to finding coefficient of x12  in (x+ x+ x+...... +x9)6

this simplifies to x12 in [(1-x10)/(1-x)]6                        sum of GP series

= coefficient of x12   in (1-x10)6 . (1-x)-6                      

=coefficient of x12 in   [(1-x)-6  - 6C1 x10 (1-x)-6 + ...............]                          Binomial expansion

=12+6-1C6-1    -  6.  [coefficient of x2 in (1-x)-6]

=17C5 - 6. 7C5

=6188 - 126

=6062 

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