The number of integers which lie between 1 and ${10}^6$ and which have the sum of the digits equal to 12, is

Consider the product

$\left(x^0+x^1+x^2\dots +x^9\right)\left(x^0+x^1+x^2\dots +x^9\right)\dots 6times$

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

= Coefficient of $x^{12}$ in ${\left(x^0+x^1+x^2+\dots +x^9\right)}^6$

$\begin{array}{l}=\text{ Coeff. of }{x}^{12}\text{ in }{\left(\frac{1-x^{10}}{1-x}\right)}^6\\ =\text{ Coeff. of }{x}^{12}\text{ in }{\left(1-x^{10}\right)}^6(1-x{)}^{-6}\end{array}$

$=\text{ Coeff. of }{x}^{12}\text{ in }(1-x{)}^{-6}\left(1{-}^6{\mathrm{C}}_1{x}^{10}+\dots \right)$

$=\text{ Coeff. of }{x}^{12}\text{ in }(1-x{)}^{-6}{-}^6{C}_1\cdot \text{ Coeff. of }{x}^2\text{ in }(1-x{)}^{-6}$

${=}^{12+6-1}{C}_{6-1}{-}^6{C}_1{\times }^{2+6-1}{C}_{6-1}{=}^{17}{C}_5-6{\times }^7{C}_5$

= 6062.