Similarly, in the expansion of \((x+y)^5\text{,}\) there will be only one \(x^5\) term and one \(y^5\) term. This is because to get an \(x^5\text{,}\) we need to use the \(x\) term in each of the copies of the binomial \((x+y)\text{,}\) and similarly for \(y^5\text{.}\) What about \(x^4y\text{?}\) To get terms like this, we need to use four \(x\)'s and one \(y\text{,}\) so we need exactly one of the five binomials to contribute a \(y\text{.}\) There are 5 choices for this, so there are 5 ways to get \(x^4y\text{,}\) so the coefficient of \(x^4y\) is 5. This is also the coefficient for \(xy^4\) for the same (but opposite) reason: there are 5 ways to pick which of the 5 binomials contribute the single \(x\text{.}\) So far we have