Let’s solve each problem one by one, step by step.
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Problem 1: How many ways can the letters in the word MAINE be arranged if you use all 5 letters (and repetition is allowed)?
- The word “MAINE” has 5 different letters: M, A, I, N, E.
- We are using
all 5 letters, and
repetition is allowed.
- That means for each of the 5 positions in the arrangement, we can choose any of the 5 letters — even if we’ve used it before.
- So:
- Position 1: 5 choices
- Position 2: 5 choices
- Position 3: 5 choices
- Position 4: 5 choices
- Position 5: 5 choices
- Total = 5 × 5 × 5 × 5 × 5 = 5⁵
Calculate 5⁵:
5 × 5 = 25
25 × 5 = 125
125 × 5 = 625
625 × 5 =
3125
✔ Answer for Problem 1:
3125
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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 total.
- We’re making a
3-digit code, and
no repeating digits.
- But note: In codes, leading zeros are usually allowed (like "012" is a valid 3-digit code). So we don’t have to worry about excluding zero from the first position unless specified — and it isn’t here.
- So:
- First digit: 10 choices (0–9)
- Second digit: 9 remaining choices (can’t repeat)
- Third digit: 8 remaining choices
- Total = 10 × 9 × 8 =
720
✔ Answer for Problem 2:
720
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Problem 3: How many ways can 3 people win 1st, 2nd, and 3rd place?
Wait — this question doesn’t say how many people are racing! Looking back at the original image, there was an example: “There are 8 people running a race...” but this problem just says “3 people”.
Actually, re-reading: “How many ways can 3 people win 1st, 2nd, and 3rd place?”
This implies we are arranging
3 specific people into 3 distinct places.
So:
- Person A, B, C
- Who gets 1st? 3 choices
- Who gets 2nd? 2 remaining choices
- Who gets 3rd? 1 choice left
→ 3 × 2 × 1 =
6
Alternatively, since n = r = 3, we can use n! = 3! = 6.
✔ Answer for Problem 3:
6
*(Note: If the problem meant “from a larger group”, it would have said so. As written, it’s about arranging 3 people into 3 spots.)*
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Problem 4: How many ways can 8 students sit in 4 seats?
- We have 8 students, but only 4 seats.
- Order matters (who sits where), and no repetition (one student per seat).
- This is permutation without repetition: P(n, r) = n! / (n - r)!
- Here, n = 8, r = 4
→ P(8, 4) = 8! / (8 - 4)! = 8! / 4!
But we don’t need to calculate full factorials:
8! / 4! = (8 × 7 × 6 × 5 × 4!) / 4! = 8 × 7 × 6 × 5
Calculate:
8 × 7 = 56
56 × 6 = 336
336 × 5 =
1680
✔ Answer for Problem 4:
1680
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Final Answer:
1. 3125
2. 720
3. 6
4. 1680
Parent Tip: Review the logic above to help your child master the concept of permutations and combinations worksheet with answers.
