Past exam of the year 1 of the computer science course of the University of Oxford / 2015 / Discrete mathematics, continuous mathematics, and probability / 2 / b

For $0\le k\le n$, the adjacent binomial coefficients obeySince $\left(\genfrac{}{}{0ex}{}{n}{k}\right)>0$ and $k+1>0$, the sign is determined by the numerator:The printed codomain ${\mathbb{N}}_{+}$ is inconsistent with the displayed map: $g(n)=-1$, and when $n$ is odd, $g((n-1)/2)=0$. With the usual extension $\left(\genfrac{}{}{0ex}{}{n}{k}\right)=0$ for $k>n$, $g$ is a function ${\mathbb{N}}_0\to \mathbb{Z}$, and $g(k)=0$ for $k>n$; the positive range above remains unchanged. Without that extension, the natural domain is $\{0,\dots ,n\}$.^[ref]

In the intended range $0\le k\le n$, division by the positive binomial coefficient $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ givesConsequently the intended remainder isThere is a necessary qualification to the printed unrestricted domain. If binomial coefficients outside $0\le k\le n$ are extended by zero, then every $k>n$ satisfies $g(k)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)=0$, regardless of $n\mathrm{mod}3$. For example, $n=3,k=4$ satisfies the literal equality but has $n\mathrm{mod}3=0$. Thus the literal extended-domain premise alone does not determine a remainder; the boxed conclusion requires $0\le k\le n$, or equivalently here $\left(\genfrac{}{}{0ex}{}{n}{k}\right)\ne 0$.