New Year Maths $2019$
$$\color{red}{\binom {20}{19}} +\color{orange}{\binom {19}{16} +{\binom {16}{14}}}+ \color{green}{\binom {14}{12} +\binom {12}{7}}+ \color{blue}{\binom {7}{6}+ \binom {6}{5}}+ \color{purple}{\binom {5}{3}+\binom {3}{1}}+ \color{magenta}{\binom {1}{0}} =\color{red}{2019}$$
$$\color{red}{\binom{20}{19}}+\color{orange}{\binom{19}{18}}+\color{orange}{\binom{18}{17}}+\color{green}{\binom{17}{16}}+\color{green}{\binom{16}{12}}+\color{blue}{\binom{12}{11}}+\color{blue}{\binom{11}{10}}+\color{purple}{\binom{10}{8}}+\color{purple}{\binom{8}{5}}+\color{magenta}{\binom{5}{0}}\color{orange}=\color{red}{2019}$$ $\color{red}{\text{Happy New Year!}}$
A simple python script yielded the following results for a,b,c,d,e,f,g,h,k such that $19>a> b> c> \cdots > k\geq 0$
16,14,12,7,6,5,3,1,0
16,14,12,7,6,5,3,2,0
17,14,10,7,6,5,3,1,0
17,14,10,7,6,5,3,2,0
18,15,14,10,9,5,4,2,0
18,15,14,10,9,5,4,3,1
18,15,14,10,9,5,4,3,2
18,17,16,12,10,9,7,6,1
18,17,16,12,10,9,7,6,5
18,17,16,12,11,9,7,2,0
18,17,16,12,11,9,7,5,0
18,17,16,12,11,9,7,6,2
18,17,16,12,11,9,7,6,4
18,17,16,12,11,10,8,3,0
18,17,16,12,11,10,8,5,0