What is the coefficient of x^8 y^2 in the expansion of (x + y)^10?
Accepted Solution
A:
Answer:The coefficient of [tex]x^{8}y^{2}[/tex] is [tex]45[/tex].Step-by-step explanation:By the binomial theorem we have that [tex](x+y)^{n}=\sum_{k=0}^{n} {{n}\choose{k}} x^{n-k}y^{k}[/tex].where [tex]{n \choose k}[/tex] is known has a binomial coefficient and it can be compute by [tex]\dfrac{n!}{(n-k)! k!}[/tex]. The symbol [tex]n![/tex] stands for the factorial of [tex]n[/tex], that is to say, [tex]n!=1\times 2\times \cdots \times n[/tex]So if you have [tex]n=10[/tex] then we can replace on the expression of the binomial theorem to obtain[tex](x+y)^{10}=\sum_{k=0}^{10}{10\choose k}x^{10-k}y^{k}[/tex]In order to obtain the coefficient of [tex]x^{8}y^{2}[/tex] we find the term where [tex]k=2[/tex], that is to say,[tex]{10\choose 2} x^{10-2}y^{2}=\frac{10!}{(10-2)!2!}x^{8}y^{2}=45x^{8}y^{2}[/tex]. We conclude that the coefficient we were looking for is [tex]45[/tex].