Permutation and Combination quiz with 10 questions on selecting and arranging items.

Let's solve each of these Permutation and Combination problems step by step. We'll determine whether each situation involves combinations (order doesn't matter) or permutations (order matters), and apply the appropriate formulas.

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🔹 Key Formulas:

- Combination (selection):
$$
C(n, r) = \binom{n}{r} = \frac{n!}{r!(n - r)!}
$$

- Permutation (arrangement):
$$
P(n, r) = \frac{n!}{(n - r)!}
$$

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✔ Problem 1:

In how many ways can a nurse select 10 students from a class of 25 to be vaccinated against covid 19?

- This is a selection problem — order does not matter.
- Use combination:
$$
\binom{25}{10} = \frac{25!}{10!(25 - 10)!} = \frac{25!}{10! \cdot 15!}
$$
- Compute:
$$
\binom{25}{10} = 3,268,760
$$

✔ Answer: 3,268,760 ways.

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✔ Problem 2:

In how many ways can a nurse arrange 10 students, 3 at a time for vaccination?

- "Arrange" means order matters, so this is a permutation.
- Selecting and arranging 3 out of 10:
$$
P(10, 3) = \frac{10!}{(10 - 3)!} = \frac{10!}{7!} = 10 \times 9 \times 8 = 720
$$

✔ Answer: 720 ways.

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✔ Problem 3:

Petermark helped his parents on gardening; he arranged 5 varieties of snake plants (sansevieria) from tallest to shortest.

- He is arranging all 5 varieties in a specific order (from tallest to shortest).
- Since it's a complete arrangement of all items, and order matters, this is a permutation of 5 items.
- But note: The order is fixed as tallest to shortest — only one way to do this.
- So, even though it’s an arrangement, the condition "from tallest to shortest" implies only one possible arrangement.

✔ Answer: 1 way.

> 💡 This is a trick question — although it's about arrangement, the constraint makes it unique.

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✔ Problem 4:

Roy shuffled a pack of 52 playing cards and asked his friend, Gino, to choose any three cards.

- Gino is choosing 3 cards from 52 — order doesn’t matter.
- Use combination:
$$
\binom{52}{3} = \frac{52 \times 51 \times 50}{3 \times 2 \times 1} = 22,100
$$

✔ Answer: 22,100 ways.

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✔ Problem 5:

Rizza chooses a combination of four digits from 0 to 9 without repetition for her passcode.

- She picks 4 distinct digits from 0–9 → 10 digits total.
- But since it’s a passcode, order matters (e.g., 1234 ≠ 4321).
- So this is a permutation of 4 digits chosen from 10:
$$
P(10, 4) = 10 \times 9 \times 8 \times 7 = 5,040
$$

✔ Answer: 5,040 possible passcodes.

> ❗ Note: Even though it says "combination", in context of passcodes, order matters → permutation.

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✔ Problem 6:

During the final round of the 100-m dash, six students were closing in on the finish line. The top three winners were awarded with gold, silver and bronze medals.

- The top 3 positions are distinct (gold, silver, bronze), so order matters.
- Choose and arrange 3 out of 6:
$$
P(6, 3) = 6 \times 5 \times 4 = 120
$$

✔ Answer: 120 ways.

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✔ Problem 7:

Rose places two red squash, three ampalaya and ten tomatoes in a vegetable basket. Her son, Jaymar, selects three vegetables from the basket for his friend.

- Total vegetables: 2 + 3 + 10 = 15
- Jaymar selects 3 vegetables — but they are of different types, and we assume vegetables of same type are indistinct unless stated otherwise.
- However, the question asks: “selects three vegetables” — no mention of order or distinction beyond type.
- If vegetables of the same kind are indistinguishable, then we need to consider combinations with constraints based on availability.

But since it's a selection of 3 vegetables from a mixed group, and no specification about distinguishability, we assume that each vegetable is distinct (common assumption unless stated otherwise).

So, total distinct vegetables = 15
Choose 3:
$$
\binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455
$$

✔ Answer: 455 ways.

> ⚠️ If vegetables of same type are identical, we'd need to break into cases, but since not specified, standard interpretation is distinct objects.

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✔ Problem 8:

In how many ways can you select a committee of 3 students out of 10 students?

- Committee → selection, order doesn’t matter.
- Use combination:
$$
\binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120
$$

✔ Answer: 120 ways.

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✔ Problem 9:

How many 3-letter words with or without meaning can be formed out of the letters of the word SWING when repetition of letters is not allowed?

- Word: SWING → 5 distinct letters: S, W, I, N, G
- Repetition not allowed → pick 3 different letters and arrange them.
- Permutation of 3 letters from 5:
$$
P(5, 3) = 5 \times 4 \times 3 = 60
$$

✔ Answer: 60 words.

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✔ Problem 10:

A committee including 3 boys and 4 girls is to be formed from a group of 10 boys and 12 girls. How many different committees can be formed from the group?

- Need: 3 boys from 10 → $\binom{10}{3}$
- And: 4 girls from 12 → $\binom{12}{4}$
- Multiply both (independent choices):

$$
\binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120
$$
$$
\binom{12}{4} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495
$$
$$
\text{Total committees} = 120 \times 495 = 59,400
$$

✔ Answer: 59,400 different committees.

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✔ Final Answers Summary:

| Question | Answer |
|--------|--------|
| 1 | 3,268,760 |
| 2 | 720 |
| 3 | 1 |
| 4 | 22,100 |
| 5 | 5,040 |
| 6 | 120 |
| 7 | 455 |
| 8 | 120 |
| 9 | 60 |
| 10 | 59,400 |

Let me know if you’d like explanations in simpler language or visual breakdowns!