I've been considering entry i in row n of Pascal's Triangle's Triangle, so for \(\displaystyle{U}\le{i}\le{n}\), we have

\(\displaystyle{\left({n}{i}\right)}={n}\frac{!}{{{i}!{\left({n}-{i}\right)}!}}\)

\(\rightarrow\) The row of (n k) are the binomial coefficients (n k) evaluated at

\(k=0,1,2,3,4,5,6,7,8,9\)

\(\displaystyle{\left({n}{0}\right)}={n}\frac{!}{{{0}!{\left({n}-{0}\right)}!}}\)

\(\displaystyle{\left({n}{1}\right)}={n}\frac{!}{{{1}!{\left({n}-{1}\right)}!}}\)

\(\displaystyle{\left({n}{2}\right)}={n}\frac{!}{{{2}!{\left({n}-{2}\right)}!}}\)

\(\displaystyle{\left({n}{3}\right)}={n}\frac{!}{{{3}!{\left({n}-{3}\right)}!}}\)

\(\displaystyle{\left({n}{4}\right)}={n}\frac{!}{{{4}!{\left({n}-{4}\right)}!}}\)

\(\displaystyle{\left({n}{5}\right)}={n}\frac{!}{{{5}!{\left({n}-{5}\right)}!}}\)

\(\displaystyle{\left({n}{6}\right)}={n}\frac{!}{{{6}!{\left({n}-{6}\right)}!}}\)

\(\displaystyle{\left({n}{7}\right)}={n}\frac{!}{{{7}!{\left({n}-{7}\right)}!}}\)

\(\displaystyle{\left({n}{8}\right)}={n}\frac{!}{{{8}!{\left({n}-{8}\right)}!}}\)

\(\displaystyle{\left({n}{9}\right)}={n}\frac{!}{{{9}!{\left({n}-{9}\right)}!}}\)

I've been considering entry i in row n of Pascal's Triangle, so for \(\displaystyle{0}\le{i}\le{n}\), we have

\(\displaystyle{\left({n}{i}\right)}={n}\frac{!}{{{i}!{\left({n}-{i}\right)}!}}\)