Consider the following expansions of the power ${\left(x\+y\right)}^{n}$ into a sum of terms:
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${\left(x\+y\right)}^{2}equals;{x}^{2}plus;2\cdot x\cdot yplus;{y}^{2}$

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${\left(x\+y\right)}^{3}equals;{x}^{3}plus;3\cdot {x}^{2}\cdot yplus;3\cdot x\cdot {y}^{2}plus;{y}^{2}$

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${\left(x\+y\right)}^{4}equals;{x}^{4}plus;4\cdot {x}^{3}\cdot yplus;6\cdot {x}^{2}\cdot {y}^{2}plus;4\cdot x\cdot {y}^{3}plus;{y}^{4}$

This expansion can be expressed more compactly using the Binomial Formula:
${\left(x\+y\right)}^{n}equals;\sum _{kequals;0}^{n}\left(\genfrac{}{}{0ex}{}{n}{k}\right)\cdot {x}^{nk}\cdot {y}^{k}equals;\sum _{kequals;0}^{n}\left(\genfrac{}{}{0ex}{}{n}{k}\right)\cdot {x}^{k}\cdot {y}^{nk}$.
The binomial coefficient $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$, also written $C\left(n\,k\right)$ or ${}_{n}{C}_{k}$ and pronounced n choose k, is the number of ways of choosing a subset of k objects from a group of n objects. Its value can be given more explicitly as
$\left(\genfrac{}{}{0ex}{}{n}{k}\right)\=\frac{n\cdot \left(n1\right)\cdot ..\.\cdot \left(nk\+1\right)}{k\cdot \left(k1\right)\cdot ..period;\cdot 1}equals;\frac{nexcl;}{kexcl;\cdot \left(nk\right)excl;}$ for $k\le n$.
As you may have noticed, the coefficients of the expansion of the ${n}^{\mathrm{th}}$ power ${\left(x\+y\right)}^{n}$ correspond directly to the numbers in the ${n}^{\mathrm{th}}$ row of Pascal's Triangle:
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${\left(x\+y\right)}^{{2}}equals;{1}\cdot {x}^{2}plus;{2}\cdot x\cdot yplus;{1}\cdot {y}^{2}$

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${\left(x\+y\right)}^{{3}}equals;{1}\cdot {x}^{3}plus;{3}\cdot {x}^{2}\cdot yplus;{3}\cdot x\cdot {y}^{2}plus;{1}\cdot {y}^{2}$

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${\left(x\+y\right)}^{{4}}equals;{1}\cdot {x}^{4}plus;{4}\cdot {x}^{3}\cdot yplus;{6}\cdot {x}^{2}\cdot {y}^{2}plus;{4}\cdot x\cdot {y}^{3}plus;{1}\cdot {y}^{4}$ and so on...

In other words, the number in the ${k}^{\mathrm{th}}$ position of the ${n}^{\mathrm{th}}$ row of Pascal's Triangle is $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$. For example, the number in position 0 of row 1 is $\left(\genfrac{}{}{0ex}{}{1}{0}\right)\=1$ while the number in position 2 of row 5 is $\left(\genfrac{}{}{0ex}{}{5}{2}\right)\=10$.
Also, due to the symmetry of Pascal's triangle, we can easily see that $\left(\genfrac{}{}{0ex}{}{n}{k}\right)equals;\left(\genfrac{}{}{0ex}{}{n}{nk}\right)$. Finally, looking back upon the the original way of constructing the triangle, in which the two numbers in the row above are added together to get the current value, we see that
$\left(\genfrac{}{}{0ex}{}{n1}{k1}\right)\+\left(\genfrac{}{}{0ex}{}{n1}{k}\right)\=\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ for $n\>0$ and $0\le k\le n$.