Question

How many integers between $1$ and $1000000$ have the sum of the digits equal to $18$?

Answer 1

Integers between $1$ and $1000000$ will be $1,2,3,4,5or6$ digits.
And given the sum of digits $=18$
Thus we need to obtain the number of solutions of the equation
$x_1+x_2+x_3+x_4+x_5+x_6=18...\left(1\right)$
where $0\le x_i\le 9,i=1,2,3,4,5,6$
Therefore, the number of solutions of the equation $\left(1\right)$, will be
$=\text{Coefficient of}{x}^{18}\text{in}(x^0+x^1+x^2+x^3+...+x^9{)}^6$
$=\text{Coefficient of}{x}^{18}\text{in}{\left(\frac{1-x^{10}}{1-x}\right)}^6$
$=\text{Coefficient of}{x}^{18}\text{in}(1-x^{10}{)}^6(1-x{)}^{-6}$
$=\text{Coefficient of}{x}^{18}\text{in}(1-6x^{10})(1+{}^6{C}_{1}x+{}^7{C}_2{x}^2+...+{}^{13}{C}_8{x}^8+...+{}^{23}{C}_{18}{x}^{18}+...)$
${=}^{23}{C}_{18}-6.{}^{13}{C}_8$
${=}^{23}{C}_5-6.{}^{13}{C}_5$
$=\frac{\mathrm{23.22.21.20.19}}{\mathrm{1.2.3.4.5}}-6.\frac{\mathrm{13.12.11.10.9.}}{\mathrm{1.2.3.4.5}}$
$=33649-7722$
$=25927$