In a contest in which 8 contestants are entered, in how many ways can the 5 distinct prizes be awarded?

permutations:

nPr = n(n - 1)(n - 2) ... (n - r + 1) = n! / (n - r)!

6720

Well, it all depends on whether we have a fair contest or not. If the contest is rigged, then I can give you a precise answer – 10,000 ways. Why? Because the person in charge of the contest already decided who is going to win each prize. But if we assume a fair contest, where every contestant has an equal chance of winning, then we have 8 choices for the first prize, 7 choices for the second prize (since one contestant already won the first prize), 6 choices for the third prize, 5 choices for the fourth prize, and 4 choices for the fifth prize. So, we multiply these possibilities together and get a total of 8 x 7 x 6 x 5 x 4 = 6,720 ways to award the prizes in a fair and unbiased contest. Happy prize-giving!

To determine the number of ways the 5 distinct prizes can be awarded among the 8 contestants, we can use the concept of combinations.

Since each prize is distinct, we can assign one prize to each of the 5 contestants in 8 different ways.

For the first prize, we have 8 choices, for the second prize, we have 7 choices, for the third prize, we have 6 choices, for the fourth prize, we have 5 choices, and for the fifth prize, we have 4 choices.

Therefore, the total number of ways to award the prizes is:

8 * 7 * 6 * 5 * 4 = 8! / (8 - 5)! = 8! / 3! = 8 * 7 * 6 = 336.

In this case, 336 different combinations of awarding the prizes can be made among the 8 contestants.

n = 8

r = 5

nPr = 8!/(8-5)! = ?

8! = "8 factorial" = 8 * 7 * 6* 5 * 4 * 3 * 2 * 1

I hope this helps a little more.