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

Let's solve each problem step-by-step using the Fundamental Counting Principle, which states:

> If one event can occur in $ m $ ways and a second event can occur independently in $ n $ ways, then the two events together can occur in $ m \times n $ ways.

We'll apply this principle to all problems. Let’s go through them one by one.

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1. 3 digits followed by 4 letters

#### (a) Digits and letters can be repeated
- Digits: 0–9 → 10 choices per digit
- Letters: A–Z → 26 choices per letter

So:
- 3 digits: $ 10 \times 10 \times 10 = 10^3 = 1000 $
- 4 letters: $ 26 \times 26 \times 26 \times 26 = 26^4 = 456,976 $

Total passwords:
$$
1000 \times 456,976 = \boxed{456,976,000}
$$

#### (b) Digits and letters cannot be repeated
- Digits: First digit: 10 choices, second: 9, third: 8 → $ 10 \times 9 \times 8 = 720 $
- Letters: First: 26, second: 25, third: 24, fourth: 23 → $ 26 \times 25 \times 24 \times 23 = 358,800 $

Total passwords:
$$
720 \times 358,800 = \boxed{258,336,000}
$$

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2. 2 digits followed by 5 letters

#### (a) Repeated allowed
- 2 digits: $ 10^2 = 100 $
- 5 letters: $ 26^5 = 11,881,376 $

Total:
$$
100 \times 11,881,376 = \boxed{1,188,137,600}
$$

#### (b) No repetition
- Digits: $ 10 \times 9 = 90 $
- Letters: $ 26 \times 25 \times 24 \times 23 \times 22 = 7,893,600 $

Total:
$$
90 \times 7,893,600 = \boxed{710,424,000}
$$

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3. Suits: Jacket, Shirt, Tie, Pants

- Jackets: 3
- Shirts: 6
- Ties: 8
- Pants: 4

Total combinations:
$$
3 \times 6 \times 8 \times 4 = \boxed{576}
$$

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4. Baseball batting order – 9 players, pitcher bats last

- Pitcher is fixed in the last position → only 1 choice for that spot
- Remaining 8 players can be arranged in any order: $ 8! $

$$
8! = 40,320
$$

So total batting orders: $ \boxed{40,320} $

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5. Eight-digit numbers: leading digit ≠ 0, last digit ≠ 1

We are forming 8-digit numbers with constraints:

- Digits: 0–9
- Leading digit (first): cannot be 0 → 9 choices (1–9)
- Last digit: cannot be 1 → 9 choices (0,2–9)
- Middle 6 digits: no restrictions → 10 choices each

But we must be careful: the first and last digits may overlap (e.g., if first digit is 1, last digit can't be 1 — so we need to avoid double-counting).

Let’s break it down:

Let’s denote positions:
$ d_1 d_2 d_3 d_4 d_5 d_6 d_7 d_8 $

Constraints:
- $ d_1 \neq 0 $ → 9 choices
- $ d_8 \neq 1 $ → 9 choices
- $ d_2 $ to $ d_7 $: 10 choices each

Now, are these independent? Yes, because even though both $ d_1 $ and $ d_8 $ have restrictions, they are different positions, and there's no restriction on overlapping values unless specified.

So:
- $ d_1 $: 9 choices (1–9)
- $ d_2 $ to $ d_7 $: $ 10^6 $
- $ d_8 $: 9 choices (0,2–9)

Total:
$$
9 \times 10^6 \times 9 = \boxed{81,000,000}
$$

✔ This is correct because the constraints are independent.

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6. 4-digit odd numbers, no repeated digits

A 4-digit number: $ d_1d_2d_3d_4 $

Conditions:
- $ d_1 \neq 0 $
- Number is odd → $ d_4 $ must be odd: {1,3,5,7,9} → 5 choices
- No digit repeats

We need to count valid numbers where:
- $ d_4 \in \{1,3,5,7,9\} $
- $ d_1 \neq 0 $
- All digits distinct

Best approach: Fix the last digit (odd), then choose the rest.

Let’s consider cases based on whether the last digit is 1 or not, since 1 is special (it could affect $ d_1 $).

But better: Use case analysis based on the last digit, but keep track of digit availability.

Let’s do it step-by-step.

Step 1: Choose $ d_4 $: must be odd → 5 choices: {1,3,5,7,9}

For each such choice, we’ll count how many ways to choose $ d_1,d_2,d_3 $, with:
- $ d_1 \neq 0 $
- No digit repeats

We can use total permutations minus invalid ones, but let's proceed carefully.

Better: Choose $ d_4 $ first, then choose $ d_1 $, then $ d_2 $, then $ d_3 $, subtracting used digits.

Let’s fix $ d_4 $. There are 5 odd digits. But depending on whether $ d_4 = 1 $ or not, the number of available digits for $ d_1 $ changes slightly.

Let’s split into two cases:

#### Case 1: $ d_4 = 1 $

- Then digit 1 is used.
- $ d_1 $: cannot be 0 or 1 → choices: 2–9 → 8 choices
- $ d_2 $: 8 remaining digits (10 total − 2 used: 1 and $ d_1 $)
- $ d_3 $: 7 remaining

So:
- $ d_1 $: 8
- $ d_2 $: 8
- $ d_3 $: 7

Total for this case: $ 8 \times 8 \times 7 = 448 $

Wait — but actually, after choosing $ d_1 $, we have 8 digits left (since 2 used), then $ d_2 $: 8 choices, $ d_3 $: 7.

Yes.

So: $ 8 \times 8 \times 7 = 448 $

But wait: $ d_1 $ has 8 choices (2–9), then $ d_2 $: 8 digits left (excluding $ d_1 $ and 1), $ d_3 $: 7.

Yes.

So total for $ d_4 = 1 $: $ 448 $

#### Case 2: $ d_4 \in \{3,5,7,9\} $ → 4 choices

Each of these is an odd digit ≠ 1.

Say $ d_4 = x $, where $ x \in \{3,5,7,9\} $

Then:
- Digit $ x $ is used
- $ d_1 $: cannot be 0 or $ x $ → choices: 1–9 except $ x $ → 8 choices (since 9 digits from 1–9, minus 1)
- $ d_2 $: 8 digits left (10 − 2 used: $ x $ and $ d_1 $)
- $ d_3 $: 7 digits left

So for each such $ d_4 $, number of options: $ 8 \times 8 \times 7 = 448 $

There are 4 such values for $ d_4 $

So total: $ 4 \times 448 = 1,792 $

Now add both cases:
- $ d_4 = 1 $: 448
- $ d_4 = 3,5,7,9 $: 1,792

Total 4-digit odd numbers with no repeated digits:
$$
448 + 1,792 = \boxed{2,240}
$$

✔ Answer: $ \boxed{2,240} $

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7. Alaska license plate: 3 letters followed by 3 digits

#### (a) Repeats allowed
- Letters: $ 26^3 = 17,576 $
- Digits: $ 10^3 = 1,000 $
- Total: $ 17,576 \times 1,000 = \boxed{17,576,000} $

#### (b) No repeats
- Letters: $ 26 \times 25 \times 24 = 15,600 $
- Digits: $ 10 \times 9 \times 8 = 720 $
- Total: $ 15,600 \times 720 = \boxed{11,232,000} $

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8. Roll a die: less than 3, then even, then odd

Rolls happen sequentially.

- Less than 3: possible rolls: 1 or 2 → 2 outcomes
- Even: 2, 4, 6 → 3 outcomes
- Odd: 1, 3, 5 → 3 outcomes

Since these are independent rolls (assuming three separate rolls), total ways:
$$
2 \times 3 \times 3 = \boxed{18}
$$

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9. Prime, then 6

Prime numbers on a die: 2, 3, 5 → 3 primes

Then roll a 6 → 1 outcome

So:
$$
3 \times 1 = \boxed{3}
$$

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10. Red and blue die: sum of 6

Possible pairs $(r,b)$ such that $ r + b = 6 $

List them:
- (1,5)
- (2,4)
- (3,3)
- (4,2)
- (5,1)

(6,0) invalid

So 5 outcomes.

Answer: $ \boxed{5} $

Note: Since dice are distinguishable (red and blue), order matters. So (1,5) ≠ (5,1). We already listed them separately.

✔ Yes, 5 ways.

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11. Pick a face card, then a spade (with replacement)

Standard deck: 52 cards

Face cards: Jack, Queen, King → 3 per suit → 4 suits → $ 3 \times 4 = 12 $ face cards

Spades: 13 cards

With replacement → probabilities independent

Number of ways:
- First pick: 12 choices (face cards)
- Second pick: 13 choices (spades)

Total:
$$
12 \times 13 = \boxed{156}
$$

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12. Pick a king, then a heart, then a five (with replacement)

Again, with replacement → independent picks.

- Kings: 4 (one per suit)
- Hearts: 13
- Fives: 4 (one per suit)

So:
- First: 4 choices (king)
- Second: 13 choices (heart)
- Third: 4 choices (five)

Total:
$$
4 \times 13 \times 4 = \boxed{208}
$$

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## ✔ Final Answers Summary:

1. (a) $ 456,976,000 $, (b) $ 258,336,000 $
2. (a) $ 1,188,137,600 $, (b) $ 710,424,000 $
3. $ 576 $
4. $ 40,320 $
5. $ 81,000,000 $
6. $ 2,240 $
7. (a) $ 17,576,000 $, (b) $ 11,232,000 $
8. $ 18 $
9. $ 3 $
10. $ 5 $
11. $ 156 $
12. $ 208 $

Let me know if you want explanations in a more visual format or need help with a specific problem!