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Question # What is the row of Pascal’s triangle containing the binomial coefficients (nk),0≤k≤9?

Binomial probability
ANSWERED What is the row of Pascal’s triangle containing the binomial coefficients $$\displaystyle{\left({n}{k}\right)},{0}≤{k}≤{9}?$$ 2021-01-23

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