There are ten contestants in a chess tournament. The number of ways in
which the first, second, and third place prizes can be awarded is closest to?
To find the number of ways the first, second, and third place prizes can be awarded among ten contestants, we can use the concept of permutations.
In a permutation, the order of the elements matters. In this case, the order of the contestants matters because it determines who receives the first, second, and third place prizes.
To calculate the number of ways to arrange the contestants, we start with the first place. There are ten contestants, so the first place can be one of the ten contestants.
For the second place, there are nine remaining contestants since one has already been chosen for the first place. So, the second place can be one of the nine contestants.
Similarly, for the third place, there are eight remaining contestants since two have already been chosen for the first and second places.
Therefore, the total number of ways the first, second, and third place prizes can be awarded is:
10 * 9 * 8 = 720
Hence, the closest value to the number of ways is 720.