Properties of Euler’s Numbers (Combination Numbers)

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Euler’s symbol $\left({\begin{array}{c}n\\k\end{array}}\right)$ , read $n$ over $k$ , is used in combinatorial theory very often. So it is useful to get know several important properties of these so-called combination numbers.

Symmetry

$\left({\begin{array}{c}n\\k\end{array}}\right)=\left({\begin{array}{c}n\\n-k\end{array}}\right)$ Proof of symmetry:

${\frac {n\,!}{k\,!(n-k)\,!}}={\frac {n\,!}{(n-k)\,!(n-(n-k))\,!}}$ Specific cases

{\begin{aligned}\left({\begin{array}{c}n\\0\end{array}}\right)&=&{\frac {n\,!}{0\,!(n-0)\,!}}=1\\&&\\\left({\begin{array}{c}n\\1\end{array}}\right)&=&{\frac {n\,!}{1\,!(n-1)\,!}}=n\\&&\\\left({\begin{array}{c}0\\0\end{array}}\right)&=&\left({\begin{array}{c}n\\n\end{array}}\right)=1\\&&\\\left({\begin{array}{c}n\\k\end{array}}\right)&=&0\ {\text{for}}\ k>n\geq 0\end{aligned}} Sum of two Euler’s numbers

$\left({\begin{array}{c}n\\k\end{array}}\right)+\left({\begin{array}{c}n\\k+1\end{array}}\right)=\left({\begin{array}{c}n+1\\k+1\end{array}}\right)$ Derivation of the property:

{\begin{aligned}{\frac {n\,!}{k\,!(n-k)\,!}}+{\frac {n\,!}{(k+1)\,!(n-(k+1))\,!}}&=&{\frac {(k+1)n\,!}{(k+1)k\,!(n-k)\,!}}+{\frac {(n-k)n\,!}{(k+1)\,!(n-(k+1))\,!(n-k)}}\\&&\\&=&{\frac {n\,!((k+1)+(n-k))}{(k+1)\,!(n-k)\,!}}\\&&\\&=&{\frac {n\,!(n+1)}{((n+1)-(k+1))\,!(k+1)\,!}}\\&&\\&=&{\frac {(n+1)\,!}{((n+1)-(k+1))\,!(k+1)\,!}}\\&&\\&=&\left({\begin{array}{c}n+1\\k+1\end{array}}\right)\end{aligned}} Euler’s numbers and binomial coefficients

The following table contains in the left column an expression in the form $(a+b)^{n}$ and in the right column summands obtained by expansion of the expression in the left column.

Pascal’s triangle

In Pascal triangle, there are all the coefficients from the above presented table. Notice additive dependence between two rows of the triangle.

$(a+b)^{6}=a^{6}+6a^{5}b+15a^{4}b^{2}+20a^{3}b^{3}+15a^{2}b^{4}+6ab^{5}+b^{6}$ Binomial theorem

Binomial theorem documents the mentioned dependence between Euler’s numbers and combination numbers. $(a+b)^{n}=\left({\begin{array}{c}n\\0\end{array}}\right)a^{n}+\left({\begin{array}{c}n\\1\end{array}}\right)a^{n-1}b+\left({\begin{array}{c}n\\2\end{array}}\right)a^{n-2}b^{2}+\dots +\left({\begin{array}{c}n\\n-1\end{array}}\right)ab^{n-1}+\left({\begin{array}{c}n\\n\end{array}}\right)b^{n}=$ $=\sum _{k=0}^{n}\left({\begin{array}{c}n\\k\end{array}}\right)a^{n-k}b^{k}$ 