Permutations and Combinations Practice 2 worksheet featuring ten problems on permutations and combinations.

Permutations and Combinations Practice 2

Here is the solution to each problem, along with explanations:

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#### 1. Permutation or Combination?
How many ways 5 runners can be arranged for 1st, 2nd, and 3rd place

- Type: Permutation (order matters)
- Reasoning: The positions (1st, 2nd, 3rd) are distinct, so the order in which the runners finish matters.
- Calculation:
\[
P(5, 3) = \frac{5!}{(5-3)!} = \frac{5!}{2!} = 5 \times 4 \times 3 = 60
\]
- Answer: Permutation, 60 ways

---

#### 2. Permutation or Combination?
Selecting 3 types of fruit from a basket of 10 different types of fruit

- Type: Combination (order does not matter)
- Reasoning: The selection of fruit types does not depend on the order in which they are chosen.
- Calculation:
\[
C(10, 3) = \binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10!}{3! \cdot 7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120
\]
- Answer: Combination, 120 ways

---

#### 3. Permutation or Combination?
Choosing 4 books from a bin of 20 books

- Type: Combination (order does not matter)
- Reasoning: The selection of books does not depend on the order in which they are chosen.
- Calculation:
\[
C(20, 4) = \binom{20}{4} = \frac{20!}{4!(20-4)!} = \frac{20!}{4! \cdot 16!} = \frac{20 \times 19 \times 18 \times 17}{4 \times 3 \times 2 \times 1} = 4845
\]
- Answer: Combination, 4845 ways

---

#### 4. Permutation or Combination?
The batting order for a softball team of 20 players

- Type: Permutation (order matters)
- Reasoning: The batting order is a specific arrangement of players, so the order matters.
- Calculation:
\[
P(20, 9) = \frac{20!}{(20-9)!} = \frac{20!}{11!} = 20 \times 19 \times 18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12
\]
This is a large number, but the exact value is not necessary for identifying the type.
- Answer: Permutation

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#### 5. How many ways can you arrange these 7 numbers: 1, 2, 3, 5, 7, 8, 9? (Don't repeat.)

- Type: Permutation (all numbers are distinct, and order matters)
- Reasoning: We are arranging all 7 numbers, so we use the factorial of 7.
- Calculation:
\[
7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040
\]
- Answer: 5040 ways

---

#### 6. You are choosing 4 pizza toppings from a menu of 10 toppings. How many different pizzas can you make? (You can repeat toppings.)

- Type: Combination with repetition (multiset combination)
- Reasoning: Since toppings can be repeated, we use the formula for combinations with repetition:
\[
C(n + r - 1, r) = \binom{n + r - 1}{r}
\]
where \( n = 10 \) (types of toppings) and \( r = 4 \) (number of toppings to choose).
- Calculation:
\[
C(10 + 4 - 1, 4) = \binom{13}{4} = \frac{13!}{4!(13-4)!} = \frac{13!}{4! \cdot 9!} = \frac{13 \times 12 \times 11 \times 10}{4 \times 3 \times 2 \times 1} = 715
\]
- Answer: 715 ways

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#### 7. Joseph wants 2 different types of soda for a party. There are 10 different types of soda at the store. How many different combinations of soda could he choose?

- Type: Combination (order does not matter)
- Reasoning: The selection of soda types does not depend on the order in which they are chosen.
- Calculation:
\[
C(10, 2) = \binom{10}{2} = \frac{10!}{2!(10-2)!} = \frac{10!}{2! \cdot 8!} = \frac{10 \times 9}{2 \times 1} = 45
\]
- Answer: 45 ways

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#### 8. The student body is electing class officers for president, vice president, and secretary. If there are 300 students, how many different possible selections are there?

- Type: Permutation (order matters)
- Reasoning: The positions (president, vice president, secretary) are distinct, so the order in which the students are selected matters.
- Calculation:
\[
P(300, 3) = \frac{300!}{(300-3)!} = \frac{300!}{297!} = 300 \times 299 \times 298
\]
\[
300 \times 299 \times 298 = 26,730,600
\]
- Answer: 26,730,600 ways

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#### 9. How many different 4-letter arrangements can you make with the letters ABCDEFG? (Allow repeats.)

- Type: Permutation with repetition
- Reasoning: Each position in the 4-letter arrangement can be any of the 7 letters, and repetition is allowed.
- Calculation:
\[
7^4 = 7 \times 7 \times 7 \times 7 = 2401
\]
- Answer: 2401 ways

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#### 10. Find the number of possible outcomes from choosing 5 basketball players from a group of 30 that tried out.

- Type: Combination (order does not matter)
- Reasoning: The selection of players does not depend on the order in which they are chosen.
- Calculation:
\[
C(30, 5) = \binom{30}{5} = \frac{30!}{5!(30-5)!} = \frac{30!}{5! \cdot 25!} = \frac{30 \times 29 \times 28 \times 27 \times 26}{5 \times 4 \times 3 \times 2 \times 1} = 142,506
\]
- Answer: 142,506 ways

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Final Answers:

\[
\boxed{
\begin{array}{ll}
1. & \text{Permutation, 60 ways} \\
2. & \text{Combination, 120 ways} \\
3. & \text{Combination, 4845 ways} \\
4. & \text{Permutation} \\
5. & 5040 \text{ ways} \\
6. & 715 \text{ ways} \\
7. & 45 \text{ ways} \\
8. & 26,730,600 \text{ ways} \\
9. & 2401 \text{ ways} \\
10. & 142,506 \text{ ways} \\
\end{array}
}
\]