First, second and third prizes are to be awarded at an engineering fair in which 13 different exhibits have been entered. In how many different ways can these three distinct prizes be awarded to the exhibits?

Introduction / Context:
This question is about assigning distinct prizes to different exhibits. Because first, second and third prizes are different in prestige, the order in which exhibits receive these prizes matters. Therefore we are dealing with permutations, not combinations. The set of exhibits is fixed and has 13 distinct entries, and we must determine how many ways we can pick and order three of them to receive the prizes.

Given Data / Assumptions:

• Total exhibits entered in the fair = 13.
• There are three distinct prizes: first, second and third.
• Each exhibit can receive at most one prize.
• The same exhibit cannot receive more than one position.

Concept / Approach:
We are choosing and ordering 3 exhibits from a set of 13 distinct exhibits. The number of such ordered selections is given by permutations, written as P(13, 3). This can be calculated as 13 * 12 * 11, representing the number of ways to assign the three positions sequentially from the pool of exhibits.

Step-by-Step Solution:
Step 1: Choose an exhibit for the first prize. There are 13 possible exhibits that can receive first prize. Step 2: Choose an exhibit for the second prize. After the first prize is awarded, only 12 exhibits remain. Step 3: Choose an exhibit for the third prize. Now 11 exhibits remain eligible for third prize. Step 4: Multiply the number of choices for each prize. Total number of prize assignments = 13 * 12 * 11. Compute: 13 * 12 = 156; then 156 * 11 = 1716.

Verification / Alternative check:
Using the permutation formula, we have P(13, 3) = 13! / (13 - 3)! = 13! / 10!. This equals 13 * 12 * 11, which matches the step by step multiplication above. Both approaches give 1716, confirming the correctness of the result.

Why Other Options Are Wrong:
The value 1736 differs from 1716 by 20 and does not correspond to any standard permutation calculation. The values 1216 and 1346 are also arbitrary and are not equal to 13 * 12 * 11. The option 2197 equals 13^3, which would be relevant if we allowed repetition of exhibits across prizes, but that is not allowed here because each exhibit can receive at most one prize.

Common Pitfalls:
A common mistake is to use combinations instead of permutations, calculating C(13, 3) instead of P(13, 3). Combinations ignore the order of selection, whereas here the order (first, second, third) is crucial. Another pitfall is to miscalculate the product 13 * 12 * 11 or to allow an exhibit to win multiple prizes. Keeping the distinction between ordered and unordered selections clear avoids these errors.

Final Answer:
The three distinct prizes can be awarded in 1716 different ways.