At an annual flower show, 6 different entries are to be arranged in a row.
a) How many different arrangements of the entries are possible?
b) If the owners of the 1st, 2nd, and 3rd place entries will be awarded prizes of $100, $50, and $25 respectively, how many ways can the prizes be awarded?
This is a duplicate post of the same question. See answer at your other (later) post
120
To answer these questions, we can use the concept of permutations.
a) To determine the number of different arrangements of the entries, we need to find the number of permutations of the 6 entries in a row. Since each entry is unique, we can use the formula for permutations of distinct objects, which is n!. In this case, n is 6. Thus, the number of different arrangements is 6! (read as "6 factorial").
To calculate 6!, we multiply all the positive integers less than or equal to 6 together: 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720. Therefore, there are 720 different arrangements of the entries.
b) To determine the number of ways the prizes can be awarded, we need to consider the number of permutations of the top 3 placements. Since there are 6 entries, there are 6 possible choices for the first-place winner, 5 possible choices for the second-place winner (as one entry has already been chosen), and 4 possible choices for the third-place winner (as two entries have already been chosen).
Using the formula for permutations of distinct objects, the number of ways the prizes can be awarded is given by 6P3, which can be found using the formula n! / (n - r)!, where n is the total number of objects and r is the number of objects being chosen. In this case, n is 6 and r is 3.
Calculating 6P3, we find: 6P3 = 6! / (6 - 3)! = 6! / 3! = (6 x 5 x 4 x 3 x 2 x 1) / (3 x 2 x 1) = 120. Therefore, there are 120 ways to award the prizes.