Fundamental Counting Principles Worksheet | Study Guides, Projects ... - Free Printable

Problem: Fundamental Counting Principles Worksheet

We will solve each problem step by step using 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.

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#### Problem 1:
Determine how many different computer passwords are possible if:
- (a) digits and letters can be repeated.
- (b) digits and letters cannot be repeated.
- 3 digits followed by 4 letters.

##### Solution:
- (a) Digits and letters can be repeated:
- There are 10 possible digits (0–9) for each of the 3 digit positions.
- There are 26 possible letters (A–Z) for each of the 4 letter positions.
- Total number of passwords:
\[
10^3 \times 26^4 = 1000 \times 456976 = 456,976,000
\]

- Answer for (a): \( \boxed{456,976,000} \)

- (b) Digits and letters cannot be repeated:
- For the 3 digits: The first digit has 10 choices, the second has 9 choices, and the third has 8 choices.
- For the 4 letters: The first letter has 26 choices, the second has 25 choices, the third has 24 choices, and the fourth has 23 choices.
- Total number of passwords:
\[
10 \times 9 \times 8 \times 26 \times 25 \times 24 \times 23 = 10 \times 9 \times 8 \times 26 \times 25 \times 24 \times 23 = 78,624,000
\]

- Answer for (b): \( \boxed{78,624,000} \)

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#### Problem 2:
Determine how many different computer passwords are possible if:
- (a) digits and letters can be repeated.
- (b) digits and letters cannot be repeated.
- 2 digits followed by 5 letters.

##### Solution:
- (a) Digits and letters can be repeated:
- There are 10 possible digits (0–9) for each of the 2 digit positions.
- There are 26 possible letters (A–Z) for each of the 5 letter positions.
- Total number of passwords:
\[
10^2 \times 26^5 = 100 \times 11,881,376 = 1,188,137,600
\]

- Answer for (a): \( \boxed{1,188,137,600} \)

- (b) Digits and letters cannot be repeated:
- For the 2 digits: The first digit has 10 choices, and the second has 9 choices.
- For the 5 letters: The first letter has 26 choices, the second has 25 choices, the third has 24 choices, the fourth has 23 choices, and the fifth has 22 choices.
- Total number of passwords:
\[
10 \times 9 \times 26 \times 25 \times 24 \times 23 \times 22 = 78,936,000
\]

- Answer for (b): \( \boxed{78,936,000} \)

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#### Problem 3:
A men's department store sells 3 different suit jackets, 6 different shirts, 8 different ties, and 4 different pairs of pants. How many different suits consisting of a jacket, shirt, tie, and pants are possible?

##### Solution:
- There are 3 choices for the jacket, 6 choices for the shirt, 8 choices for the tie, and 4 choices for the pants.
- Total number of different suits:
\[
3 \times 6 \times 8 \times 4 = 576
\]

- Answer: \( \boxed{576} \)

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#### Problem 4:
A baseball manager is determining the batting order for the team. The team has 9 players, but the manager definitely wants the pitcher to bat last. How many batting orders are possible?

##### Solution:
- Since the pitcher must bat last, we only need to arrange the remaining 8 players in the first 8 positions.
- The number of ways to arrange 8 players is given by \( 8! \):
\[
8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40,320
\]

- Answer: \( \boxed{40,320} \)

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#### Problem 5:
How many eight-digit numbers can be formed if the leading digit cannot be a zero and the last digit cannot be 1?

##### Solution:
- The leading digit has 9 choices (1–9).
- Each of the middle 6 digits has 10 choices (0–9).
- The last digit has 9 choices (0–9, excluding 1).
- Total number of eight-digit numbers:
\[
9 \times 10^6 \times 9 = 9 \times 1,000,000 \times 9 = 81,000,000
\]

- Answer: \( \boxed{81,000,000} \)

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#### Problem 6:
How many 4-digit odd numbers can be formed if no digit can be repeated?

##### Solution:
- An odd number must end in an odd digit. There are 5 odd digits: 1, 3, 5, 7, 9.
- Choose the last digit (odd digit): 5 choices.
- Choose the first digit (cannot be 0 or the last digit): 8 choices (1–9, excluding the last digit).
- Choose the second digit (cannot be the first or last digit): 8 choices (0–9, excluding the first and last digits).
- Choose the third digit (cannot be the first, second, or last digit): 7 choices.
- Total number of 4-digit odd numbers:
\[
5 \times 8 \times 8 \times 7 = 2,240
\]

- Answer: \( \boxed{2,240} \)

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#### Problem 7:
The standard configuration for an Alaska license plate is 3 letters followed by 3 digits. How many different license plates are possible if:
- (a) Letters and digits can be repeated.
- (b) Letters and digits cannot be repeated.

##### Solution:
- (a) Letters and digits can be repeated:
- There are 26 possible letters for each of the 3 letter positions.
- There are 10 possible digits for each of the 3 digit positions.
- Total number of license plates:
\[
26^3 \times 10^3 = 17,576 \times 1,000 = 17,576,000
\]

- Answer for (a): \( \boxed{17,576,000} \)

- (b) Letters and digits cannot be repeated:
- For the 3 letters: The first letter has 26 choices, the second has 25 choices, and the third has 24 choices.
- For the 3 digits: The first digit has 10 choices, the second has 9 choices, and the third has 8 choices.
- Total number of license plates:
\[
26 \times 25 \times 24 \times 10 \times 9 \times 8 = 11,232,000
\]

- Answer for (b): \( \boxed{11,232,000} \)

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#### Problem 8:
A single die is rolled. How many ways can you roll a number less than 3, then an even number, and then an odd number?

##### Solution:
- Roll a number less than 3: Possible outcomes are 1 and 2 (2 choices).
- Roll an even number: Possible outcomes are 2, 4, and 6 (3 choices).
- Roll an odd number: Possible outcomes are 1, 3, and 5 (3 choices).
- Total number of ways:
\[
2 \times 3 \times 3 = 18
\]

- Answer: \( \boxed{18} \)

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#### Problem 9:
A single die is rolled. How many ways can you roll a number that is prime, followed by a 6?

##### Solution:
- Prime numbers on a die are 2, 3, and 5 (3 choices).
- The second roll must be a 6 (1 choice).
- Total number of ways:
\[
3 \times 1 = 3
\]

- Answer: \( \boxed{3} \)

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#### Problem 10:
A red die and a blue die are rolled. In how many ways can you get a sum of 6?

##### Solution:
- Possible pairs of rolls that sum to 6:
- (1, 5)
- (2, 4)
- (3, 3)
- (4, 2)
- (5, 1)
- Total number of ways:
\[
5
\]

- Answer: \( \boxed{5} \)

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#### Problem 11:
From a standard deck of cards, how many ways can you pick a face card, then a spade? Provided you put the first card back.

##### Solution:
- A standard deck has 52 cards.
- Face cards are Jacks, Queens, and Kings, and there are 3 face cards per suit, so there are \( 3 \times 4 = 12 \) face cards.
- Spades are one of the 4 suits, so there are 13 spades.
- Since the first card is put back, the total number of ways:
\[
12 \times 13 = 156
\]

- Answer: \( \boxed{156} \)

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#### Problem 12:
From a standard deck of cards, how many ways can you pick a king, a heart, and then a five? Provided you put the cards back into the deck.

##### Solution:
- A standard deck has 52 cards.
- Kings: There are 4 kings in the deck.
- Hearts: There are 13 hearts in the deck.
- Fives: There are 4 fives in the deck.
- Since the cards are put back, the total number of ways:
\[
4 \times 13 \times 4 = 208
\]

- Answer: \( \boxed{208} \)

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Final Answers:

1. (a) \( \boxed{456,976,000} \), (b) \( \boxed{78,624,000} \)
2. (a) \( \boxed{1,188,137,600} \), (b) \( \boxed{78,936,000} \)
3. \( \boxed{576} \)
4. \( \boxed{40,320} \)
5. \( \boxed{81,000,000} \)
6. \( \boxed{2,240} \)
7. (a) \( \boxed{17,576,000} \), (b) \( \boxed{11,232,000} \)
8. \( \boxed{18} \)
9. \( \boxed{3} \)
10. \( \boxed{5} \)
11. \( \boxed{156} \)
12. \( \boxed{208} \)