Question

What is the coefficient of x^8 y^2 in the expansion of (x + y)^10?

Answer 1

Answer:

The coefficient of $x^{8}y^{2}$ is $45$.

Step-by-step explanation:

By the binomial theorem we have that

$(x+y)^{n}=\sum_{k=0}^{n} {{n}\choose{k}} x^{n-k}y^{k}$.

where ${n \choose k}$ is known has a binomial coefficient and it can be compute by $\dfrac{n!}{(n-k)! k!}$. The symbol $n!$ stands for the factorial of $n$, that is to say,  $n!=1\times 2\times \cdots \times n$

So if you have $n=10$ then we can replace on the expression of the binomial theorem to obtain

$(x+y)^{10}=\sum_{k=0}^{10}{10\choose k}x^{10-k}y^{k}$

In order to obtain the coefficient of $x^{8}y^{2}$ we find the term where $k=2$, that is to say,

${10\choose 2} x^{10-2}y^{2}=\frac{10!}{(10-2)!2!}x^{8}y^{2}=45x^{8}y^{2}$. We conclude that the coefficient we were looking for is $45$.