worksheet a2 fundamental counting principle Doc Template | pdfFiller - Free Printable

Problem: Solve the tasks from the worksheet "Fundamental Counting Principle, Factorials, Permutations Intro."

#### Task 1:
A restaurant offers four sizes of pizza, two types of crust, and eight toppings. How many possible combinations of pizza with one topping are there?

Solution:
To solve this, we use the Fundamental Counting Principle, which states that if there are \( m \) ways to do one thing and \( n \) ways to do another, then there are \( m \times n \) ways to do both.

- Sizes of pizza: 4 options
- Types of crust: 2 options
- Toppings: 8 options (since only one topping is allowed per pizza)

The total number of combinations is:
\[
4 \text{ (sizes)} \times 2 \text{ (crusts)} \times 8 \text{ (toppings)} = 64
\]

Answer:
\[
\boxed{64}
\]

---

#### Task 2:
How many ways can 5 paintings be lined up on a wall?

Solution:
This is a permutation problem because the order in which the paintings are arranged matters. The number of ways to arrange \( n \) distinct items is given by \( n! \) (n factorial), where:
\[
n! = n \times (n-1) \times (n-2) \times \cdots \times 1
\]

Here, \( n = 5 \):
\[
5! = 5 \times 4 \times 3 \times 2 \times 1 = 120
\]

Answer:
\[
\boxed{120}
\]

---

#### Task 3:
Rob has 4 shirts, 3 pairs of pants, and 2 pairs of shoes that all coordinate. How many outfits can he put together?

Solution:
Again, we use the Fundamental Counting Principle. Each outfit consists of one shirt, one pair of pants, and one pair of shoes.

- Shirts: 4 options
- Pants: 3 options
- Shoes: 2 options

The total number of outfits is:
\[
4 \text{ (shirts)} \times 3 \text{ (pants)} \times 2 \text{ (shoes)} = 24
\]

Answer:
\[
\boxed{24}
\]

---

#### Task 4:
Grace loves to eat salad! How many salads can she put together if she can pick out one dressing from 4 choices, one vegetable from 5 choices, and one fruit from 7 choices?

Solution:
This is another application of the Fundamental Counting Principle. Each salad consists of one dressing, one vegetable, and one fruit.

- Dressings: 4 options
- Vegetables: 5 options
- Fruits: 7 options

The total number of salads is:
\[
4 \text{ (dressings)} \times 5 \text{ (vegetables)} \times 7 \text{ (fruits)} = 140
\]

Answer:
\[
\boxed{140}
\]

---

#### Task 5a:
PA license plates have 3 letters followed by 4 numbers. If the same letter or number can be repeated, how many can be made?

Solution:
Each position in the license plate can be filled independently. There are 26 letters in the alphabet and 10 digits (0–9).

- Letters: 3 positions, each with 26 choices.
- Numbers: 4 positions, each with 10 choices.

The total number of license plates is:
\[
26 \times 26 \times 26 \times 10 \times 10 \times 10 \times 10 = 26^3 \times 10^4
\]

Calculating:
\[
26^3 = 26 \times 26 \times 26 = 17576
\]
\[
10^4 = 10000
\]
\[
26^3 \times 10^4 = 17576 \times 10000 = 175760000
\]

Answer:
\[
\boxed{175760000}
\]

---

#### Task 5b:
If the same letter or number CANNOT be repeated, how many can be made?

Solution:
Now, each position must have a unique letter or number.

- Letters: 3 positions, with decreasing choices (26, 25, 24).
- Numbers: 4 positions, with decreasing choices (10, 9, 8, 7).

The total number of license plates is:
\[
26 \times 25 \times 24 \times 10 \times 9 \times 8 \times 7
\]

Calculating step-by-step:
\[
26 \times 25 = 650
\]
\[
650 \times 24 = 15600
\]
\[
10 \times 9 = 90
\]
\[
90 \times 8 = 720
\]
\[
720 \times 7 = 5040
\]
\[
15600 \times 5040 = 78624000
\]

Answer:
\[
\boxed{78624000}
\]

---

#### Task 6:
How many 5-digit numbers can be formed using 0–9?

Solution:
Each digit in the 5-digit number can be any of the 10 digits (0–9). Since repetition is allowed, each of the 5 positions has 10 choices.

The total number of 5-digit numbers is:
\[
10 \times 10 \times 10 \times 10 \times 10 = 10^5
\]

Calculating:
\[
10^5 = 100000
\]

Answer:
\[
\boxed{100000}
\]

---

#### Task 7:
How many 5-digit numbers can be formed if each one uses all the digits 0, 1, 2, 3, 4 without repetition?

Solution:
This is a permutation problem because the digits must be used exactly once, and the order matters. We need to find the number of permutations of 5 distinct digits.

The number of permutations of 5 distinct digits is:
\[
5! = 5 \times 4 \times 3 \times 2 \times 1 = 120
\]

However, we must exclude numbers that start with 0 (since they would not be 5-digit numbers).

- Total permutations: \( 5! = 120 \)
- Numbers starting with 0: Treat "0" as fixed in the first position, and permute the remaining 4 digits:
\[
4! = 4 \times 3 \times 2 \times 1 = 24
\]

Subtract the invalid numbers:
\[
120 - 24 = 96
\]

Answer:
\[
\boxed{96}
\]

---

#### Task 8:
In how many ways can 6 bicycles be parked in a row?

Solution:
This is a permutation problem because the order in which the bicycles are parked matters. The number of ways to arrange \( n \) distinct items is \( n! \).

Here, \( n = 6 \):
\[
6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720
\]

Answer:
\[
\boxed{720}
\]

---

#### Task 9:
Evaluate (show all your work):
a. \( 6! \)
b. \( 9! \)

Solution:
a. Calculate \( 6! \):
\[
6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720
\]

b. Calculate \( 9! \):
\[
9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1
\]
Breaking it down:
\[
9 \times 8 = 72
\]
\[
72 \times 7 = 504
\]
\[
504 \times 6 = 3024
\]
\[
3024 \times 5 = 15120
\]
\[
15120 \times 4 = 60480
\]
\[
60480 \times 3 = 181440
\]
\[
181440 \times 2 = 362880
\]
\[
362880 \times 1 = 362880
\]

Answers:
\[
\boxed{720} \quad \text{(for } 6!\text{)}
\]
\[
\boxed{362880} \quad \text{(for } 9!\text{)}
\]

---

Final Answers:

1. \(\boxed{64}\)
2. \(\boxed{120}\)
3. \(\boxed{24}\)
4. \(\boxed{140}\)
5a. \(\boxed{175760000}\)
5b. \(\boxed{78624000}\)
6. \(\boxed{100000}\)
7. \(\boxed{96}\)
8. \(\boxed{720}\)
9a. \(\boxed{720}\)
9b. \(\boxed{362880}\)