Math Problem Statement

Identify the following as either a Permutation or a Combination. Then answer the question. Show Your Work on your worksheet when solving. Do not just put into your calculator.

A) There are 13 applications for 4 Computer Programmer Positions. How many ways can these be selected?

Combination

Permutation

   ways.

B) A group of 54 people are going to run a race. The top three runners earn gold, silver, and bronze medals, respectively.

CombinationPermutation

   ways the medals could be earned.

Solution

Let's address each part of the question:

Part A

Problem: There are 13 applications for 4 Computer Programmer Positions. How many ways can these be selected?

Type: Permutation

Reasoning: The order matters because different arrangements of the 4 positions would constitute different outcomes.

Calculation:

The number of permutations of 13 items taken 4 at a time is given by: P(13,4)=13!(134)!=13!9!P(13, 4) = \frac{13!}{(13-4)!} = \frac{13!}{9!}

This simplifies to: P(13,4)=13×12×11×10P(13, 4) = 13 \times 12 \times 11 \times 10

Calculation steps:

  1. Calculate 13×1213 \times 12: 13×12=15613 \times 12 = 156

  2. Multiply the result by 11: 156×11=1716156 \times 11 = 1716

  3. Multiply the result by 10: 1716×10=171601716 \times 10 = 17160

Therefore, there are 17,16017,160 ways to select and arrange 4 Computer Programmer Positions from 13 applications.

Part B

Problem: A group of 54 people are going to run a race. The top three runners earn gold, silver, and bronze medals, respectively.

Type: Permutation

Reasoning: The order matters because the positions are distinct (gold, silver, and bronze).

Calculation:

The number of permutations of 54 items taken 3 at a time is given by: P(54,3)=54!(543)!=54!51!P(54, 3) = \frac{54!}{(54-3)!} = \frac{54!}{51!}

This simplifies to: P(54,3)=54×53×52P(54, 3) = 54 \times 53 \times 52

Calculation steps:

  1. Calculate 54×5354 \times 53: 54×53=286254 \times 53 = 2862

  2. Multiply the result by 52: 2862×52=1488242862 \times 52 = 148824

Therefore, there are 148,824148,824 ways the medals could be earned by the top three runners.

Would you like more details or explanations on any part of the solutions?

  1. Why is Part A considered a permutation problem?
  2. Why is the order of selection important in permutation problems?
  3. Can you describe a real-life example where combinations are used instead of permutations?
  4. How would the calculations change if Part A was a combination problem?
  5. What is the difference between permutations and combinations in mathematical terms?
  6. Can you show the factorial breakdown for one of the calculations?
  7. How would you calculate the total possible permutations if an additional position is added in Part A?
  8. What would be the number of ways to distribute the medals if there were 4 instead of 3 in Part B?

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Math Problem Analysis

Mathematical Concepts

Permutations
Combinations
Factorials

Formulas

Permutations formula P(n, k) = n! / (n - k)!
Combinations formula C(n, k) = n! / (k! * (n - k)!)

Theorems

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Suitable Grade Level

High School