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Eight people are entered in a race. If there are no ties, in how many ways can the first three places come out? a.336 b.368 c.672 d.512 Please explain how to do this problem.

Respuesta :

wolf1728
The first thing to do is to calculate how many ways you can choose 3 people from a set of eight.  In order to do this, we need to use the attached formula.
(The letter 'n' stands for the entire set and 'r' stands for the number of objects we wish to choose.)
So we wish to choose 3 people ('r') form a set of 8 ('n')
combinations = n! / r! * (n - r)!
combinations = 8 ! / (3! * 5!)
combinations = 8 * 7 * 6 * 5! / (3!) * (5!)
combinations = 8 * 7 * 6 / 3 * 2
combinations = 56

Now of those 56 combinations, the 3 people can finish in 6 different ways.
For example, persons A, B and C could finish
ABC or ACB or BAC or BCA or CAB or CBA

So to get the TOTAL combinations we multiply 56 * 6 which equals
336 so the answer is (a)

As per permutation, in this race 1st, 2nd, and the 3rd places can be awarded in 336 different ways among 8 runners.

What is permutation?

The different ways of arranging the elements of a set is called permutation.

Here, in this given question 8 runners are competing.

The different ways 1st, 2nd, and 3rd place trophies be awarded among 8 runners is = ⁸P₃ = 8! / (8 - 3)! = 8! / 5!

= (8 × 7 × 6 × 5!) / 5! = 8 × 7 × 6 = 336.

Hence, in different 336 ways 8 runners can be awarded for the 1st, 2nd and 3rd places. Option A is correct.

Learn more about permutation here: brainly.com/question/23283166

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