Permutations and Combinations Worksheet for 6th - 8th Grade ... - Free Printable

Problem Analysis:

The task involves solving a series of combinatorics and permutations problems. Each question requires careful application of principles such as permutations, combinations, and the inclusion-exclusion principle. Let's solve each part step by step.

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Question 1: How many ways are there to line up Bob, Carl, and Dan at the bus office?

#### Solution:
To line up three people (Bob, Carl, and Dan), we need to find the number of permutations of these three individuals. The formula for the number of permutations of \( n \) distinct objects is \( n! \).

Here, \( n = 3 \):
\[
3! = 3 \times 2 \times 1 = 6
\]

Thus, the number of ways to line up Bob, Carl, and Dan is:
\[
\boxed{6}
\]

---

Question 2: How many different committees of 6 people can be formed from a freshman class of 20 students?

#### Solution:
This is a combination problem because the order in which the students are chosen does not matter. The formula for combinations is:
\[
\binom{n}{r} = \frac{n!}{r!(n-r)!}
\]
where \( n \) is the total number of students, and \( r \) is the number of students to be chosen.

Here, \( n = 20 \) and \( r = 6 \):
\[
\binom{20}{6} = \frac{20!}{6!(20-6)!} = \frac{20!}{6! \cdot 14!}
\]

Simplify the factorials:
\[
\binom{20}{6} = \frac{20 \times 19 \times 18 \times 17 \times 16 \times 15}{6 \times 5 \times 4 \times 3 \times 2 \times 1}
\]

Calculate the numerator and denominator:
\[
\text{Numerator: } 20 \times 19 \times 18 \times 17 \times 16 \times 15 = 2790720
\]
\[
\text{Denominator: } 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720
\]

Divide the numerator by the denominator:
\[
\binom{20}{6} = \frac{2790720}{720} = 3876
\]

Thus, the number of different committees of 6 people that can be formed is:
\[
\boxed{3876}
\]

---

Question 3: How many different sums of coins of two dimes and three nickels are possible if the order in which they are made is different from one made in the reverse order?

#### Solution:
First, calculate the total value of the coins:
- Two dimes: \( 2 \times 10 = 20 \) cents
- Three nickels: \( 3 \times 5 = 15 \) cents

The total value is:
\[
20 + 15 = 35 \text{ cents}
\]

Next, determine the number of ways to arrange the coins. There are 5 coins in total (2 dimes and 3 nickels). The number of distinct arrangements is given by the multinomial coefficient:
\[
\binom{5}{2, 3} = \frac{5!}{2! \cdot 3!}
\]

Calculate the factorials:
\[
5! = 120, \quad 2! = 2, \quad 3! = 6
\]

Substitute into the formula:
\[
\binom{5}{2, 3} = \frac{120}{2 \cdot 6} = \frac{120}{12} = 10
\]

Thus, there are 10 different sums of coins when considering the order.

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Question 4: How many different sums of coins of two dimes and three nickels are possible if the order in which they are made does not matter?

#### Solution:
If the order does not matter, we only need to consider the total value of the coins, which is fixed at 35 cents. Since the composition of the coins (2 dimes and 3 nickels) is fixed, there is only one unique sum regardless of the order.

Thus, the number of different sums is:
\[
\boxed{1}
\]

---

Question 5: How many ways can you select four letters from the word MATH?

#### Solution:
The word "MATH" consists of 4 distinct letters. To select 4 letters from these 4 distinct letters, we are essentially choosing all of them. The number of ways to do this is:
\[
\binom{4}{4} = 1
\]

Thus, the number of ways to select four letters from the word "MATH" is:
\[
\boxed{1}
\]

---

Question 6: In a restaurant, how many ways can you expect to select three different dishes from eight different dishes on the menu?

#### Solution:
This is a combination problem where we need to choose 3 dishes out of 8. The formula for combinations is:
\[
\binom{n}{r} = \frac{n!}{r!(n-r)!}
\]
Here, \( n = 8 \) and \( r = 3 \):
\[
\binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8!}{3! \cdot 5!}
\]

Simplify the factorials:
\[
\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1}
\]

Calculate the numerator and denominator:
\[
\text{Numerator: } 8 \times 7 \times 6 = 336
\]
\[
\text{Denominator: } 3 \times 2 \times 1 = 6
\]

Divide the numerator by the denominator:
\[
\binom{8}{3} = \frac{336}{6} = 56
\]

Thus, the number of ways to select three different dishes from eight is:
\[
\boxed{56}
\]

---

Question 7: Some stores have five phones and five televisions. How many different pairs of items can be purchased if no store has both a phone and a television?

#### Solution:
Since no store has both a phone and a television, we can treat the selection of a phone and a television as independent events.

- Number of ways to choose a phone: 5
- Number of ways to choose a television: 5

The total number of different pairs of items (one phone and one television) is:
\[
5 \times 5 = 25
\]

Thus, the number of different pairs of items that can be purchased is:
\[
\boxed{25}
\]

---

Question 8: You are packing up your collection of CDs. You have 12 CDs to choose from and 10 slots in your case. How many ways can you pack your case?

#### Solution:
This is a permutation problem because the order in which the CDs are packed matters. We need to choose 10 CDs out of 12 and arrange them in the 10 slots. The number of ways to do this is given by the permutation formula:
\[
P(n, r) = \frac{n!}{(n-r)!}
\]
Here, \( n = 12 \) and \( r = 10 \):
\[
P(12, 10) = \frac{12!}{(12-10)!} = \frac{12!}{2!}
\]

Simplify the factorials:
\[
12! = 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1
\]
\[
2! = 2 \times 1 = 2
\]

Cancel out the common terms in the numerator and denominator:
\[
P(12, 10) = 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3
\]

Calculate the product:
\[
P(12, 10) = 6652800
\]

Thus, the number of ways to pack the case is:
\[
\boxed{6652800}
\]

---

Question 9: How many different sums of money can be made using one Harry Potter book, one Star Wars book, and one Lord of the Rings book if each book costs a different amount of money?

#### Solution:
If each book costs a different amount of money, the possible sums depend on the combinations of books selected. Let the costs of the books be \( H \) (Harry Potter), \( S \) (Star Wars), and \( L \) (Lord of the Rings).

The possible sums are:
1. Cost of one book: \( H, S, L \) (3 sums)
2. Cost of two books: \( H + S, H + L, S + L \) (3 sums)
3. Cost of all three books: \( H + S + L \) (1 sum)

Total number of different sums:
\[
3 + 3 + 1 = 7
\]

Thus, the number of different sums of money that can be made is:
\[
\boxed{7}
\]

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Question 10: At least one of my 25 classmates has been vaccinated. In a class of 25 students, how many different sets of students could have been vaccinated?

#### Solution:
If at least one student has been vaccinated, we need to count all possible subsets of the 25 students except the empty set. The total number of subsets of a set with \( n \) elements is \( 2^n \). Here, \( n = 25 \):
\[
2^{25}
\]

However, since we exclude the empty set (no one vaccinated), we subtract 1:
\[
2^{25} - 1
\]

Calculate \( 2^{25} \):
\[
2^{25} = 33554432
\]

Thus, the number of different sets of students who could have been vaccinated is:
\[
33554432 - 1 = 33554431
\]

Thus, the number of different sets is:
\[
\boxed{33554431}
\]

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Final Answers:

1. \(\boxed{6}\)
2. \(\boxed{3876}\)
3. \(\boxed{10}\)
4. \(\boxed{1}\)
5. \(\boxed{1}\)
6. \(\boxed{56}\)
7. \(\boxed{25}\)
8. \(\boxed{6652800}\)
9. \(\boxed{7}\)
10. \(\boxed{33554431}\)