A teacher wants to award prizes for 1st, 2nd, 3rd, 4th and 5th in the class of
30. In how many ways can the prizes be awarded (assume no two students
tie)?

Respuesta :

Answer:

2,052,086,400.

Step-by-step explanation:

Permutation

Pr = n! / (n-r)!

30P5 = 30! / (30-5)

30P5 = 30x29x28x27x26x25! / 25!

(Just cancel the 25! on the numerator and denominator for easier solving.)

30P5 = 30x29x28x27x26

30P5 = 17,100,720.It is not the final answer because we can still arrange the 5 winners.

5! = 5×4×3×2×1 = 12017,100,720 × 120 = 2,052,086,400

We will see that the total number of different combinations in which the prizes can be awarded is C = 17,100,720

How to find the total number of combinations?

First, we need to find all the selections. In this case we have:

  • prize 1
  • prize 2
  • prize 3
  • prize 4
  • prize 5.

Now we need to count the number of options for each of these selections, the number of options will depend on the number of students in the class.

  • prize 1: 30 options here
  • prize 2: 29 options (because one student already won the 1st prize)
  • prize 3: 28 options
  • prize 4: 27 options
  • prize 5: 26 options.

The total number of different combinations is equal to the product between the numbers of options, we will have:

C = 30*29*28*27*26 = 17,100,720

If you want to learn more about combinations, you can read:

https://brainly.com/question/11732255

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