By the way, a very cool (to my mind) way of computing the Euler characteristic of $\mathbb{C}P^n$ is to treat it as the $n$-fold symmetric product of $\mathbb{C}P^1 = \mathbb{S}^2$ with itself. Then, it is apparently a result of MacDonald (of Atiyah and M fame) that the Euler characteristic of an $n$-fold symmetric product of a space $X$ with itself equals $\binom{\chi(X)+n-1}{\chi(X) - 1}.$`

By the way, a very cool (to my mind) way of computing the Euler characteristic of $\mathbb{C}P^n$ is to treat it as the $n$-fold symmetric product of $\mathbb{C}P^1 = \mathbb{S}^2$ with itself. Then, it is apparently a result of MacDonald (of Atiyah and M fame) that the Euler characteristic of an $n$-fold symmetric product of a space $X$ with itself equals $\binom{\chi(X)+n-1}{\chi(X) - 1}.$`

EDIT Following @Qiaochu's penetrating remark, I looked up MacDonald's paper (I was citing from secondary sources before, shame on me). The result is:

The $k$-th Betti number of the $n$th symmetric power of $X$ is the coefficient of $x^k t^n$ in $\prod_i (1- (-x)^i t)^{- (-1)^i (-B_i)}.$ where the $B_i$ are the betti numbers of $X$. So, for Euler characteristic you evaluate this product at $x=-1,$ and find the coefficient of $t^n.$