College

State if each scenario involves a permutation or a combination. Then find the number of possibilities.

1) Castel and Joe are planning trips to three countries this year. There are 7 countries they would like to visit. One trip will be one week long, another two days, and the last three days.

2) You are setting the combination on a three-digit lock. You want to use the numbers 1, 2, and 3 but don't care what order they are in.
- [tex](b) = 3 \cdot 2 \cdot 1[/tex]

3) A team of 17 softball players needs to choose three players to refill the water cooler.

4) The student body of 10 students wants to elect a president, vice president, secretary, and treasurer.

5) There are 15 applicants for four jobs: Computer Programmer, Software Tester, Manager, and Systems Engineer.

6) There are 110 people at a meeting. They each shake hands with everyone else. How many handshakes were there?
- [tex]110 \times 109 = 11,990[/tex]

7) A group of 25 people are going to run a race. The top 8 finishers advance to the finals.
- [tex]8! \text{ (25-8)}[/tex]

8) 5 out of 13 students will ride in a car instead of a van.

9) Selecting which seven players will be in the batting order on an 11-person team.

10) A group of 45 people are going to run a race. The top three runners earn gold, silver, and bronze medals.

Answer :

In this problem, we need to determine whether each scenario involves a permutation or a combination, and then calculate the number of possible outcomes.

Let's break down each scenario:

  1. Trips to Three Countries:

    • Description: Castel and Joe are planning trips to three countries out of a possible seven. The duration of each trip is different: one week-long, one two-day, and the other three-day.
    • Analysis: The order in which they visit the countries matters because the trips have different durations. Therefore, this is a permutation problem.
    • Calculation: Use the permutation formula for selecting 3 countries from 7:
      [tex]P(7, 3) = \frac{7!}{(7-3)!} = \frac{7 \times 6 \times 5}{1} = 210[/tex]

  2. Combination Lock:

    • Description: Setting a three-digit combination on a lock using the numbers 1, 2, and 3, irrespective of the order.
    • Analysis: Since the order does not matter, this is a combination problem.
    • Calculation: There is only 1 way to choose 3 numbers when order does not matter:
      [tex]C(3, 3) = \frac{3!}{3!(3-3)!} = 1[/tex] (Note that the formula given was incorrect.)

  3. Refilling the Water Cooler:

    • Description: Choosing 3 players from a team of 17 to refill the water cooler.
    • Analysis: The order of selection does not matter, making this a combination.
    • Calculation: Use the combination formula:
      [tex]C(17, 3) = \frac{17!}{3!(17-3)!} = \frac{17 \times 16 \times 15}{3 \times 2 \times 1} = 680[/tex]

  4. Electing Student Body Officers:

    • Description: Electing 4 officers with different roles from 10 students.
    • Analysis: The roles (president, vice president, etc.) give importance to order, making this a permutation.
    • Calculation: Use the permutation formula:
      [tex]P(10, 4) = \frac{10!}{(10-4)!} = 10 \times 9 \times 8 \times 7 = 5,040[/tex]

  5. Job Applications:

    • Description: Assigning 4 specific jobs to 15 applicants.
    • Analysis: Since each job is different, this is a permutation.
    • Calculation: Use the permutation formula:
      [tex]P(15, 4) = \frac{15!}{(15-4)!} = 15 \times 14 \times 13 \times 12 = 32,760[/tex]

  6. Handshakes at a Meeting:

    • Description: 110 people shake hands with each other.
    • Analysis: Order does not matter in a handshake, making this a combination.
    • Calculation: Calculate the number of handshakes:
      [tex]C(110, 2) = \frac{110 \times 109}{2} = 5,995[/tex]
    • (The multiplication in the question was incorrect.)

  7. Running a Race - Top Finishers:

    • Description: Top 8 out of 25 advance to the finals.
    • Analysis: The order does not matter for advancing, assuming they all qualify equally, thus a combination.
    • Calculation: Use the combination formula, though the one given was incorrect.

  8. Riding in a Car:

    • Description: Selecting 5 students out of 13 to ride in a car.
    • Analysis: The order does not matter, making this a combination.
    • Calculation: Use the combination formula:
      [tex]C(13, 5) = \frac{13!}{5!(13-5)!} = \frac{13 \times 12 \times 11 \times 10 \times 9}{5 \times 4 \times 3 \times 2 \times 1} = 1,287[/tex]

  9. Batting Order:

    • Description: Selecting which 7 players will be in the batting order from an 11-person team.
    • Analysis: The order in the batting lineup matters, making this a permutation.
    • Calculation: Use the permutation formula:
      [tex]P(11, 7) = \frac{11!}{(11-7)!} = 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 = 3,991,680[/tex]

  10. Race Medals:

    • Description: The top three runners earn distinct medals (gold, silver, bronze).
    • Analysis: Since the medals are different, the order matters, thus a permutation.
    • Calculation: Use the permutation formula:
      [tex]P(45, 3) = \frac{45!}{(45-3)!} = 45 \times 44 \times 43 = 85,140[/tex]

In summary:

  • Permutation scenarios occur when the order matters.
  • Combination scenarios occur when the order does not matter and the groups are only considered by composition.

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