New Year Maths $2019$

Solution 1:

$$\color{red}{\binom {20}{19}} +\color{orange}{\binom {19}{16} +{\binom {16}{14}}}+ \color{green}{\binom {14}{12} +\binom {12}{7}}+ \color{blue}{\binom {7}{6}+ \binom {6}{5}}+ \color{purple}{\binom {5}{3}+\binom {3}{1}}+ \color{magenta}{\binom {1}{0}} =\color{red}{2019}$$

Solution 2:

$$\color{red}{\binom{20}{19}}+\color{orange}{\binom{19}{18}}+\color{orange}{\binom{18}{17}}+\color{green}{\binom{17}{16}}+\color{green}{\binom{16}{12}}+\color{blue}{\binom{12}{11}}+\color{blue}{\binom{11}{10}}+\color{purple}{\binom{10}{8}}+\color{purple}{\binom{8}{5}}+\color{magenta}{\binom{5}{0}}\color{orange}=\color{red}{2019}$$ $\color{red}{\text{Happy New Year!}}$

Solution 3:

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

Solution 4:

$$\begin{align}&{7\choose H}+\color{red}{9\choose A}+\color{blue}{11\choose P}+{\color{blue}{12\choose P}}+\color{green}{13\choose Y}+\\ &{8\choose N}+\color{magenta}{16\choose E}+{15\choose W}+\\ &\color{green}{14\choose Y}+\color{magenta}{17\choose E}+\color{red}{10\choose A}+\binom{18}{R}= 2019.\end{align}$$

$\binom76+\binom85+\binom98+\binom{10}8+\binom{11}{10}+\binom{12}{10}+\binom{13}{12}+\binom{14}{12}+\binom{15}{13}+\binom{16}{14}+\binom{17}{14}+\binom{18}{15}=2019.$