# New Year Maths $2019$

$$\;\;\;\color{red}{\binom {20}{19}}\\ \color{orange}{+\binom {19}{a} +{\binom ab}}\\ \color{green}{+\binom bc +\binom cd}\\ \color{blue}{+\binom de+ \binom ef}\\ \color{purple}{+\binom fg+\binom gh}\\ \color{magenta}{+\binom hk}\\[15pt] \color{orange}=\\[10pt] \color{red}{2019}\\[15pt] \color{orange}{a,b,} \color{green}{c,d,} \color{blue}{e,f,} \color{purple}{g,h,} \color{magenta}k\color{orange}{=?}\\[20pt]$$ $$\color{red}{\text{Happy New Year!}}$$

Note - Please feel free to post any other interesting identities based on the same theme!

• i think you're missing the g,h term. Else it is a bit too easy Commented Dec 30, 2018 at 16:06
• I was missing you ! Happy New Year ! Commented Dec 30, 2018 at 16:12
• Is python allowed, or is that cheating? :P Commented Dec 30, 2018 at 16:28
• @Anvit - Good question! $i, j$ are usually used for indexing, and probably for the same reason, elements in the standard $3\times 3$ matrix are often labelled $\left[\begin{array}&a&b&c\\d&e&f\\g&h&k\end{array}\right]$. Commented Dec 30, 2018 at 17:08
• There are $19$ up-votes.... Should I? Commented Feb 14, 2019 at 1:00

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

• Nice! We can make each of JMoravitz's solutions into a tree and and then vote which tree is the most beautiful ... Commented Dec 30, 2018 at 22:46
• Very nice animation! (+1) Commented Dec 31, 2018 at 9:21

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

• Nice! Hope you don't mind some colors :D Commented Dec 31, 2018 at 11:59
• Thank you, Zacky! Commented Dec 31, 2018 at 12:27

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

• Nice and complete solution! (+1) Commented Dec 31, 2018 at 9:22

\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}

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

• Cool! ... too bad that the same letters don't go with the same numbers on top ... did you try that? Commented Dec 30, 2018 at 22:39
• Nope, first I tried $\binom7{H}+\binom8{A}+\cdots+\binom{18}{R}$, but then had to adjust (by trial and error, no coding). Commented Dec 30, 2018 at 22:54

$$\color{red}{{20 \choose 19}+{19\choose 16}}+\color{orange}{{16\choose 3}+{3\choose 1}}+\color{green}{{1\choose 2}+{2\choose 2}}+\color{blue}{{2\choose 466}+{466\choose 465}}+\color{purple}{{465\choose 466}+{466\choose 467}}=\color{magenta}{2019}$$.