Permutations and Combinations Notes and Worksheets - Lindsay Bowden - Free Printable

Let’s solve each problem step by step.

---

Problem 1: How many ways can the letters in the word MAINE be arranged if you use all 5 letters (and repetition is allowed)?

Wait — this says “repetition is allowed”, but the word MAINE has 5 *different* letters: M, A, I, N, E. If repetition is allowed, that means we can reuse letters even though they’re not repeated in the original word.

But note: The question says “use all 5 letters” — which usually means we are arranging exactly 5 positions, and since repetition is allowed, each position can be any of the 5 letters.

So:

- n = 5 (letters to choose from: M, A, I, N, E)
- r = 5 (we are filling 5 positions)
- Repetition allowed → formula: n^r

→ 5⁵ = 5 × 5 × 5 × 5 × 5

Calculate:

5 × 5 = 25
25 × 5 = 125
125 × 5 = 625
625 × 5 = 3125

✔ So answer for #1 is 3125

---

Problem 2: How many different 3-digit codes can you make with the single digit whole numbers (without repeating)?

Single digit whole numbers: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 → that’s 10 digits.

We want 3-digit codes, no repetition.

Important: In a code, leading zero is usually allowed (like 012 is a valid code). So we don’t have to worry about excluding zero from first digit unless specified. Since it doesn’t say “number” but “code”, we assume 0 can be used anywhere.

So:

- n = 10 (digits 0–9)
- r = 3 (positions in code)
- No repetition → formula: P(n,r) = n! / (n - r)!

→ P(10,3) = 10 × 9 × 8 = ?

10 × 9 = 90
90 × 8 = 720

✔ Answer for #2 is 720

---

Problem 3: How many ways can 3 people win 1st, 2nd, and 3rd place?

This implies we are choosing and ordering 3 winners out of some group — but wait, how many total people?

Looking back at the example above: “There are 8 people running a race...” — but Problem 3 doesn’t specify.

Wait — let’s read carefully: “How many ways can 3 people win 1st, 2nd, and 3rd place?”

It doesn’t say from how many people. But in context, maybe it’s assuming we’re selecting 3 distinct people and assigning them ranks — so we need to know the total number of candidates.

Actually, looking again — perhaps it’s implied that there are only 3 people? That wouldn’t make sense because then it would just be 3! = 6.

Alternatively, maybe it’s a general case? But that doesn’t fit.

Wait — look at the structure. In the table above, one example was: “8 people running a race... 1st, 2nd, 3rd” → that used P(8,3).

But here, Problem 3 says: “How many ways can 3 people win 1st, 2nd, and 3rd place?”

This is ambiguous — but likely, it means: if there are 3 people total, and we assign them to 1st, 2nd, 3rd — then it’s just permutations of 3 people.

That makes sense as a simple case.

If there are 3 people, and we assign each to a unique rank (no ties), then:

Number of ways = 3! = 3 × 2 × 1 = 6

Alternatively, if it meant “from a larger group”, it should specify. Since it doesn’t, and given the simplicity, I think it’s assuming 3 people competing for 3 places.

Also, note: in the “Repetition Not Allowed” section, it says: “If n and r are the same number, you can use: n!”

Here, if n=3, r=3 → 3! = 6.

✔ So answer for #3 is 6

*(Note: If the problem intended a different total number of people, it’s missing info — but based on wording, 3 people total is most logical.)*

---

Problem 4: How many ways can 8 students sit in 4 seats?

We have 8 students, and only 4 seats. We are choosing 4 students out of 8 and arranging them in order (since seats are distinct — e.g., seat 1, seat 2, etc.).

So:

- n = 8 (students to choose from)
- r = 4 (seats to fill)
- No repetition (one student per seat) → formula: P(8,4) = 8! / (8-4)! = 8! / 4!

Which is: 8 × 7 × 6 × 5

Calculate:

8 × 7 = 56
56 × 6 = 336
336 × 5 = 1680

✔ Answer for #4 is 1680

---

Final Answers:

1. 3125
2. 720
3. 6
4. 1680

──────────────────────────────────────

Final Answer:
1. 3125
2. 720
3. 6
4. 1680