Pascal's coefficient identity

Using the binomial theorem $\binom{n}{k}=\frac{n!}{(n-k)! \hspace{.15cm} k!}$ is another way to find coefficients of an specific row.

We know that $(x+y)^n = \binom{n}{0} x^n y^0 + \binom{n}{1} x^{n-1} y^1 + \binom{n}{2} x^{n-2} y^2 + \cdots + \binom{n}{n-1} x^1 y^{n-1} + \binom{n}{n} x^0 y^n$ where $\binom{n}{k}$ is a specific positive integer knwon as binomial coefficient as we already said. \par

It is the $4^\text{th}$ element in the $10^\text{th}$ row which we are looking for, therefore: $\binom{10}{4}=\frac{10!}{4!(10-4)!}$

$\binom{10}{4} = \dfrac {10×9×8×7×6×5×4×3×2×1}{4×3×2×1×6×5×4×3×2×1} = \dfrac {10×9×8×7}{4×3×2×1} = \dfrac {5040}{24} = 210$

Thus, the $4^\text{th}$ element in the $10^\text{th}$ row is equal to $\binom{10}{4}= \boxed{210}$ .

Pascal's Triangle contains the values of the binomial coefficient . Each $j^\text{th}$ element in the $i^\text{th}$ row is equal to $\binom{i}{j}$ .

Thus, the $4^\text{th}$ element in the $10^\text{th}$ row is equal to $\binom{10}{4}=\boxed{210}$ .