In a (5)-place code one color is chosen from (3) colors at each place. No two consecutive places should have the same color. How many codes are possible?

Correct answer: (48) codes

The direct answer is option C: 48 codes. For the first position, any of the 3 colors may be chosen. After that, the next position cannot repeat the previous color, so it has 2 choices. The same reasoning applies to each of the remaining three positions: each has 2 choices based on the immediately preceding position. Therefore, the total is \(3\times2\times2\times2\times2=3\times2^4=48\). Option A, 81, would correspond to a different counting situation and does not enforce the no-equal-neighbours rule. Option B, 96, is too large and introduces no valid extra choice. Option C is correct. Option D, 24, is too small; it loses one factor of 2 or otherwise undercounts the four positions after the first. Notice that a color may reappear after one or more positions; only consecutive equality is forbidden. For example, red-blue-red is allowed. The exam cue is: first position has all choices, and every later position has total colors minus one choice.