New Year Maths $2019$

Question

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

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 — Anvit, Dec 30 '18 at 16:06
Are they all natural numbers? And is $19\geq a\geq b\geq c \geq\cdots\geq k$? — 2'5 9'2, Dec 30 '18 at 16:16
@alex.jordan Ideally yes. But I look forward to being surprised! — Hypergeometricx, Dec 30 '18 at 16:17
@Anvit Anything goes! Or, shall we say, a computational approach is acceptable :). — Hypergeometricx, Dec 30 '18 at 16:31
$i$ and $j$ are often unused to avoid confusing one for the other. They are also commonly used as loop variables in programming... — Daniel Mathias, Dec 30 '18 at 16:56
@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 $\left[\begin{array}&a&b&c\d&e&f\g&h&k\end{array}\right]$. — Hypergeometricx, Dec 30 '18 at 17:08
Is $9$ the least numbers of unknowns , in order to solve it? Assuming that you start again with $\binom{20}{19}$ . — dmtri, Dec 31 '18 at 08:48
Tell us your thoughts on the problem. What have you tried? Show your work. And lose the colors. — bof, Jan 01 '19 at 08:11

caverac · Answer 1 · 2018-12-30T22:34:05.440

32

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

edited Dec 30 '18 at 22:34

answered Dec 30 '18 at 21:16

caverac

19,345

3

Nice! We can make each of JMoravitz's solutions into a tree and and then vote which tree is the most beautiful ... – Bram28 Dec 30 '18 at 22:46
3

Very nice animation! (+1) – Hypergeometricx Dec 31 '18 at 09:21

score 14 · Answer 2 · edited Dec 31 '18 at 11:58

14

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

edited Dec 31 '18 at 11:58

Zacky

27,674

answered Dec 30 '18 at 17:11

Michael Rozenberg

194,933

Nice! Hope you don't mind some colors :D – Zacky Dec 31 '18 at 11:59
2

Thank you, Zacky! – Michael Rozenberg Dec 31 '18 at 12:27

JMoravitz · Accepted Answer · 2018-12-30T17:39:39.107

12

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

edited Dec 30 '18 at 17:39

answered Dec 30 '18 at 17:34

JMoravitz

79,518

Nice and complete solution! (+1) – Hypergeometricx Dec 31 '18 at 09:22

score 10 · Answer 4 · answered Dec 30 '18 at 22:33

10

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

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

answered Dec 30 '18 at 22:33

farruhota

31,482

Cool! ... too bad that the same letters don't go with the same numbers on top ... did you try that? – Bram28 Dec 30 '18 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). – farruhota Dec 30 '18 at 22:54

score 2 · Answer 5 · 2019-01-01T06:11:15.863

2

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

edited Jan 01 '19 at 06:11

answered Dec 30 '18 at 17:49

New Year Maths $2019$

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