The number of integers which lie between 1 and 10 6 and which have the sum of the digits equal to 12 is-
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Consider the product (x 0 + x 1 + x 2 +….+ x 9 ) (x 0 + x 1 + x 2 + ….. + x 9 ) … 6 factors. The number of ways in which the sum of the digits will be equal to 12 is equal to the coefficient of x 12 in the above product. So, required number of ways
= coeff. of x 12 in (x 0 + x 1 + x 2 + …. + x 9 ) 6
= coeff. of x 12 in 
= coeff. of x 12 in (1 – x 10 ) 6 (1 – x) –6 = coeff. of x
12 in (1 – x) –6 (1 – 6 C 1 x 10 +….)
= coeff. of x 12 in (1 – x) –6 – 6 C 1 . coeff. of x 2 in (1 – x) –6 =
12 + 6 – 1 C 6 – 1 – 6 C 1 × 2 + 6 – 1 C 6 – 1 = 17 C 5 – 6 × 7 C 5 = 6062.
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