Question

PART D:
#16-#20
Solve the following permutation and combination problems. Show all your work.




Permutation :

Combination :

Answer 1

Sure! I'll outline some example problems involving permutations and combinations, along with their solutions.

### Part D

#### Problem #16: Permutation
How many different ways can 3 students be selected from a group of 5 students to compete in a quiz, where the order of selection matters?

**Solution:**
The formula for permutations is given by:
\[
P(n, r) = \frac{n!}{(n - r)!}
\]
where:
- \(n\) is the total number of items (in this case, students),
- \(r\) is the number of items to arrange.

Here, \(n = 5\) and \(r = 3\).

Plugging in the values, we get:
\[
P(5, 3) = \frac{5!}{(5-3)!} = \frac{5!}{2!}
\]
Calculating the factorials:
\[
5! = 5 \times 4 \times 3 \times 2 \times 1 = 120
\]
\[
2! = 2 \times 1 = 2
\]

Now substituting back:
\[
P(5, 3) = \frac{120}{2} = 60
\]

Thus, there are **60 different ways** to arrange 3 students from a group of 5.

---

#### Problem #17: Combination
In how many different ways can a committee of 4 members be formed from a group of 10 people, where the order of selection does not matter?

**Solution:**
The formula for combinations is given by:
\[
C(n, r) = \frac{n!}{r!(n - r)!}
\]
where:
- \(n\) is the total number of items (people),
- \(r\) is the number of items to choose.

Here, \(n = 10\) and \(r = 4\).

Plugging in the values, we get:
\[
C(10, 4) = \frac{10!}{4!(10 - 4)!} = \frac{10!}{4!6!}
\]

Calculating the factorials:
\[
10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1
\]
However, since we're looking for \( \frac{10!}{4!6!} \), we can simplify:
\[
C(10, 4) = \frac{10 \times 9 \times 8 \times 7}{4!}
\]

Calculating \(4!\):
\[
4! = 4 \times 3 \times 2 \times 1 = 24
\]

Thus, we have:
\[
C(10, 4) = \frac{10 \times 9 \times 8 \times 7}{24} = \frac{5040}{24} = 210
\]

Thus, there are **210 different ways** to form a committee of 4 members from a group of 10 people.

---

#### Problem #18: Permutation
How many different ways can 4 books be arranged on a shelf if there are 6 books to choose from?

**Solution:**
Using the permutation formula:
\[
P(n, r) = \frac{n!}{(n - r)!}
\]
Here, \(n = 6\) and \(r = 4\).

Calculating:
\[
P(6, 4) = \frac{6!}{(6 - 4)!} = \frac{6!}{2!}
\]

Calculating the factorials:
\[
6! = 720 \quad \text{and} \quad 2! = 2
\]

Putting it all together:
\[
P(6, 4) = \frac{720}{2} = 360
\]

Thus, there are **360 different ways** to arrange 4 books from 6 on a shelf.

---

#### Problem #19: Combination
A group consists of 12 people. How many ways can a team of 5 be selected from this group?

**Solution:**
Using the combination formula:
\[
C(n, r) = \frac{n!}{r!(n - r)!}
\]
Where \(n = 12\) and \(r = 5\).

Calculating:
\[
C(12, 5) = \frac{12!}{5!7!}
\]

Calculating only the necessary factorials:
\[
C(12, 5) = \frac{12 \times 11 \times 10 \times 9 \times 8}{5!}
\]
And \(5! = 120\).

Thus:
\[
C(12, 5) = \frac{95040}{120} = 792
\]

So, there are **792 ways** to select a team of 5 from a group of 12 people.

---

#### Problem #20: Permutation
In how many different ways can 4 different letters be arranged if selected from a total of 8 different letters?

**Solution:**
Using the permutation formula:
\[
P(n, r) = \frac{n!}{(n - r)!}
\]
Where \(n = 8\) and \(r = 4\).

Calculating:
\[
P(8, 4) = \frac{8!}{(8 - 4)!} = \frac{8!}{4!}
\]

Calculating the necessary factorials:
\[
8! = 40320 \quad \text{and} \quad 4! = 24
\]

Thus:
\[
P(8, 4) = \frac{40320}{24} = 1680
\]

Therefore, there are **1680 different ways** to arrange 4 letters selected from 8 different letters.

---

This concludes the permutation and combination problems with their solutions. If you need further assistance or more problems, feel free to ask!