Chapter 12 - Counting and Probability - Section 12.2 Permutations and Combinations - 12.2 Assess Your Understanding - Page 875: 38

$360$

If we want to choose $k$ elements out of $n$ regarding the order, not allowing repetition, we can do this in $_{n}P_k=\frac{n!}{(n-k)!}$ ways. The order matters here when choosing the codes, thus we have to use permutations. Thus $_{6}P_{4}=\frac{6!}{(6-4)!}=6\cdot5\cdot4\cdot3=360$