I am really stuck on this one. Let $Y=\mathbb{P}^n$ be the complex projective space and let $\tilde Y$ be the blow-up of $Y$ along a linear subvariety $X$ of codimension $d$. We get the following blow-up diagram: $$\begin{matrix} E & \xrightarrow{\;j\;} & \tilde{Y} \\ \hphantom{\scriptstyle g}\downarrow {\scriptstyle g} && \hphantom{\scriptstyle f}\downarrow {\scriptstyle f} \\ X &\xrightarrow{\;i\;} & Y \end{matrix}$$ Denote by $P$ the proper transform, under $f$, of a hyperplane in $Y$.

I am trying to calculate $c_2(\tilde Y)$, and with the help of a post here at MO, I thought that I had figured it out. However, I wanted to do a quick check if nothing went wrong, but something did go wrong. Assume $n=4$ and $d=2$, then we can use the formula from Fulton's book in Example 15.4.3 to get $$c_2(\tilde Y)=f^\ast c_2(Y) - j_\ast g^\ast c_1(X) - E^2.$$ By the answer to my question by Johannes Nordström, we can write $j_\ast g^\ast c_1(X)=3(E^2 + EP)$ and this yields $$c_2(\tilde Y)=10P^2 - 3EP - 4E^2.$$ Now, we also know from the same example in Fulton's book that $c_1(\tilde Y)=f^\ast c_1(Y) - E=5P-E$. Since the blow-up map $f$ is finite of degree one, the degrees of $c_1^2(\tilde Y)c_2(\tilde Y)$ and $c_1^2(Y)c_2(Y)$ should coincide. Because I was unsure of the calculation, I asked a second question and obtained the answer that $$P^{n-b} E^b = (-1)^{b-1+n-d} \cdot \binom{b-1}{n-d}$$ Now, we can put this all together and obtain $$\begin{align*} c_1^2(\tilde Y)c_2(\tilde Y) &= (5P-E)^2(10P^2 - 3EP - 4E^2) \\&= 250P^4 - 175P^3E - 60P^2E + 37PE^3 - 4E^4 \\&= 250 + 37 + 12 = 299, \end{align*}$$ but $c_1^2(Y)c_2(Y)=(4+1)^2\cdot\frac{4(4+1)}{2} = 250$.

I do not know where the mistake is, since I find both of the answers I received very convincing, but I cannot find a flaw in my calculation either, nor do I doubt Fulton.

I am really stuck on this one. Let $Y=\mathbb{P}^n$ be the complex projective space and let $\tilde Y$ be the blow-up of $Y$ along a linear subvariety $X$ of codimension $d$. We get the following blow-up diagram: $$\begin{matrix} E & \xrightarrow{\;j\;} & \tilde{Y} \\ \hphantom{\scriptstyle g}\downarrow {\scriptstyle g} && \hphantom{\scriptstyle f}\downarrow {\scriptstyle f} \\ X &\xrightarrow{\;i\;} & Y \end{matrix}$$ Denote by $P$ the proper transform, under $f$, of a hyperplane in $Y$.

I am trying to calculate $c_2(\tilde Y)$, and with the help of a post here at MO, I thought that I had figured it out. However, I wanted to do a quick check if nothing went wrong, but something did go wrong. Assume $n=4$ and $d=2$, then we can use the formula from Fulton's book in Example 15.4.3 to get $$c_2(\tilde Y)=f^\ast c_2(Y) - j_\ast g^\ast c_1(X) - E^2.$$ By the answer to my question by Johannes Nordström, we can write $j_\ast g^\ast c_1(X)=3(E^2 + EP)$ and this yields $$c_2(\tilde Y)=10P^2 - 3EP - 4E^2.$$ Now, we also know from the same example in Fulton's book that $c_1(\tilde Y)=f^\ast c_1(Y) - E=5P-E$. Since the blow-up map $f$ is finite of degree one, the degrees of $c_1^2(\tilde Y)c_2(\tilde Y)$ and $c_1^2(Y)c_2(Y)$ should coincide. Because I was unsure of the calculation, I asked a second question and obtained the answer that $$P^{n-b} E^b = (-1)^{b-1+n-d} \cdot \binom{b-1}{n-d}$$ Now, we can put this all together and obtain $$\begin{align*} c_1^2(\tilde Y)c_2(\tilde Y) &= (5P-E)^2(10P^2 - 3EP - 4E^2) \\&= 250P^4 - 175P^3E - 60P^2E + 37PE^3 - 4E^4 \\&= 250 + 37 + 12 = 299, \end{align*}$$ but $c_1^2(Y)c_2(Y)=(4+1)^2\cdot\frac{4(4+1)}{2} = 250$.

I do not know where the mistake is, since I find both of the answers I received very convincing, but I cannot find a flaw in my calculation either, nor do I doubt Fulton.