diff --git a/_psets/4.md b/_psets/4.md
index feeec7d22e4cab20451f1530e407deb8051a4449..8fadaed01bb3b893f4b6e2f5f71c0028c56314b1 100644
--- a/_psets/4.md
+++ b/_psets/4.md
@@ -5,10 +5,39 @@ title: Problem Set 4
 ## (6.1)
 
 {:.question}
-Prove the BAC–CAB rule $$A \times (B \times C) = B(A \cdot B) - C (A \cdot B)$$ by writing it out in
+Prove the BAC–CAB rule $$A \times (B \times C) = B(A \cdot C) - C (A \cdot B)$$ by writing it out in
 the summations convention, and use it to show that $$\nabla \times (\nabla \times E) = \nabla
 (\nabla \cdot E) - \nabla^2 E$$.
 
+I'm going to use the [Levi-Civita symbol](https://en.wikipedia.org/wiki/Levi-Civita_symbol). I'm
+also going to use $$i_+$$ to represent $$(i + 1) \text{ mod } 3$$ for any coordinate index $$i$$. So
+if $$i$$ is 2, then $$i_+ = 3$$ and $$i_{+ +} = 1$$.
+
+The $$i$$th coordinate of $$A \times (B \times C)$$ is
+
+$$
+\begin{align*}
+\epsilon_{i j k} A_j \epsilon_{k l m} B_l C_m
+&= A_{i_+} \epsilon_{i_{+ +} l m} B_l C_m - A_{i_{+ +}} \epsilon_{i_+ l m} B_l C_m \\
+&= A_{i_+} (B_i C_{i_+} - B_{i_+} C_i) - A_{i_{+ +}} (B_{i_{+ +}} C_i - B_i C_{i_{+ +}}) \\
+&= A_i B_i C_i + A_{i_+} B_i C_{i_+} + A_{i_{+ +}} B_i C_{i_{+ +}} - \\
+&\phantom{x=} A_i B_i C_i - A_{i_+} B_{i_+} C_i - A_{i_{+ +}} B_{i_{+ +}} C_i  \\
+&= B_i A_j C_j - C_i A_j B_j
+\end{align*}
+$$
+
+which is the $$i$$th coordinate of $$B(A \cdot C) - C (A \cdot B)$$.
+
+If we take $$\nabla$$ to be a three-vector of partial differentiation operators, then this shows
+that
+
+$$
+\begin{align*}
+\nabla \times (\nabla \times E)
+&= \nabla (\nabla \cdot E) - E (\nabla \cdot \nabla) \\
+&= \nabla (\nabla \cdot E) - \nabla^2 E
+\end{align*}
+$$
 
 ## (6.2)