The number of integers between 1 and 1000000 that have the sum of the digits 18, is

Any number between 1 and 1000000 must be of less than seven digits. Therefore, it must be of the form a₁ a₂ a₃ a₄ a₅ a₆

where a₁, a₂, a₃, a₄, a₅, a₆ ∈ {0, 1, 2, …, 9}

According to question, sum of the digits = 18

Thus, a₁ + a₂ + a₃ + a₄ + a₅ + a₆ = 18

where 0 ≤ a_i ≤ 9, i = 1, 2, 3, …, 9.

Required number

$\begin{array}{l}=\text{ coefficient of }{\mathrm{x}}^{18}\text{ in }{\left(1+\mathrm{x}+{\mathrm{x}}^2+\dots +{\mathrm{x}}^9\right)}^6\\ =\text{ coefficient of }{\mathrm{x}}^{18}\text{ in }{\left(\frac{1-{\mathrm{x}}^{10}}{1-\mathrm{x}}\right)}^6\\ \begin{array}{l}=\text{ coefficient of }{\mathrm{x}}^{18}\text{ in }\left[{\left(1-{\mathrm{x}}^{10}\right)}^6(1-\mathrm{x}{)}^{-6}\right]\\ =\text{ coefficient of }{\mathrm{x}}^{18}\text{ in }\left[\left(1{-}^6{\mathrm{C}}_1{\mathrm{x}}^{10}\right)(1-\mathrm{x}{)}^{-6}\right]\end{array}\end{array}$

[leaving terms containing powers of x greater than 18]

$$\begin{gathered} =\text{ coefficient of }{\mathrm{x}}^{18}\text{ in }(1-\mathrm{x}{)}^{-6}{-}^6{\mathrm{C}}_1\text{. coefficient of }{\mathrm{x}}^8\text{ in }(1-\mathrm{x}{)}^{-6} \\ \begin{array}{l}{=}^{6+18-1}{\mathrm{C}}_{18}-6{\cdot }^{6+8-1}{\mathrm{C}}_{18}{=}^{23}{\mathrm{C}}_5-6{\cdot }^{13}{\mathrm{C}}_8\\ =\frac{23\cdot 22\cdot 21\cdot 20\cdot 19}{120}-6\cdot \frac{13\cdot 12\cdot 11\cdot 10\cdot 9}{120}\\ =33649-7722=25927\end{array} \end{gathered}$$