The number of permutations \(= P(4,3) = \dfrac{4!}{(4-3)!} = \dfrac{24}{1} = 24\). This counts ordered arrangements. The number of combinations would be \(C(4,3)=4\), but permutations account for all orderings of each combination (each set of 3 has \(3! = 6\) arrangements, so \(4 \times 6 = 24\)).