Permutations and Combinations Notes and Worksheets - Lindsay Bowden - Free Printable
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Step-by-step solution for: Permutations and Combinations Notes and Worksheets - Lindsay Bowden
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Show Answer Key & Explanations
Step-by-step solution for: Permutations and Combinations Notes and Worksheets - Lindsay Bowden
Let’s solve each problem 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)?
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 any letter as many times as we want when making arrangements of 5 letters.
But note: The question says “use all 5 letters” — which might mean we are arranging exactly 5 positions, and since repetition is allowed, each position can be any of the 5 letters.
So:
- n = number of things to choose from = 5 letters (M, A, I, N, E)
- r = how many things you choose = 5 positions
- Repetition allowed → formula: n^r
So:
5⁵ = 5 × 5 × 5 × 5 × 5 = 3125
✔ But wait — let’s double-check the wording: “if you use all 5 letters”. That could be misinterpreted. In permutations with repetition allowed, “using all 5 letters” doesn’t mean each letter must appear once — it just means we’re filling 5 slots, and we can repeat. Since the problem explicitly says “repetition is allowed”, we go with n^r.
Answer for #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 through 9 → that’s 10 digits.
We want 3-digit codes, no repetition.
Important: Can a code start with 0? Yes — because it’s a “code”, not necessarily a number. So 012 is a valid 3-digit code.
So:
- n = 10 (digits 0–9)
- r = 3 (positions in the code)
- No repetition → formula: n! / (n - r)!
So:
₁₀P₃ = 10! / (10 - 3)! = 10! / 7! = 10 × 9 × 8 = 720
✔ Check: First digit: 10 choices, second: 9 left, third: 8 left → 10×9×8=720
Answer for #2: 720
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Problem 3: How many ways can 3 people win 1st, 2nd, and 3rd place?
This implies there are more than 3 people total? Wait — actually, the problem doesn’t specify how many people are competing. But looking back at the example above: “There are 8 people running a race...” — so maybe this is continuing that context?
Wait — Problem 3 stands alone: “How many ways can 3 people win 1st, 2nd, and 3rd place?”
That phrasing is ambiguous. Does it mean:
A) There are only 3 people, and they are being assigned to 1st, 2nd, 3rd? → Then it’s 3! = 6
OR
B) There are more people (like 8 from previous example), and we’re choosing 3 winners out of them?
Looking at the worksheet structure — Problem 3 comes after Problem 2, and before Problem 4 (“8 students sit in 4 seats”). Also, the example above the table used 8 people for 1st/2nd/3rd.
But Problem 3 doesn’t mention a total number. Hmm.
Wait — perhaps it’s implying: “If there are 3 people, how many ways can they be ordered into 1st, 2nd, 3rd?” That would be 3!.
Alternatively, maybe it’s a trick — if only 3 people are running, then yes, 3! = 6.
But let’s look at the pattern:
- Example in table: 8 people → 1st, 2nd, 3rd → uses permutation without repetition: ₈P₃
- Problem 4: 8 students in 4 seats → ₈P₄
So likely, Problem 3 is meant to be similar — but it doesn’t say how many total people.
Wait — re-read: “How many ways can 3 people win 1st, 2nd, and 3rd place?”
This probably means: There are 3 distinct people, and we assign them to 3 distinct places → order matters → 3! = 6
Because if it were selecting 3 winners from a larger group, it would say “from X people”.
Also, in the absence of a specified total, the most logical interpretation is that we have exactly 3 people to assign to 3 positions.
So:
n = 3, r = 3, no repetition → 3! = 6
✔ Answer for #3: 6
*(Note: If the problem intended a larger pool, it should have specified. Given the wording, 6 is correct.)*
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Problem 4: How many ways can 8 students sit in 4 seats?
This is classic permutation without repetition.
- n = 8 students
- r = 4 seats
- Order matters (who sits where)
- No repetition (one student per seat)
Formula: n! / (n - r)! = ₈P₄
Calculate:
₈P₄ = 8 × 7 × 6 × 5 = ?
8 × 7 = 56
56 × 6 = 336
336 × 5 = 1680
✔ Answer for #4: 1680
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Now, let’s compile final answers.
Final Answer:
1. 3125
2. 720
3. 6
4. 1680
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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)?
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 any letter as many times as we want when making arrangements of 5 letters.
But note: The question says “use all 5 letters” — which might mean we are arranging exactly 5 positions, and since repetition is allowed, each position can be any of the 5 letters.
So:
- n = number of things to choose from = 5 letters (M, A, I, N, E)
- r = how many things you choose = 5 positions
- Repetition allowed → formula: n^r
So:
5⁵ = 5 × 5 × 5 × 5 × 5 = 3125
✔ But wait — let’s double-check the wording: “if you use all 5 letters”. That could be misinterpreted. In permutations with repetition allowed, “using all 5 letters” doesn’t mean each letter must appear once — it just means we’re filling 5 slots, and we can repeat. Since the problem explicitly says “repetition is allowed”, we go with n^r.
Answer for #1: 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 through 9 → that’s 10 digits.
We want 3-digit codes, no repetition.
Important: Can a code start with 0? Yes — because it’s a “code”, not necessarily a number. So 012 is a valid 3-digit code.
So:
- n = 10 (digits 0–9)
- r = 3 (positions in the code)
- No repetition → formula: n! / (n - r)!
So:
₁₀P₃ = 10! / (10 - 3)! = 10! / 7! = 10 × 9 × 8 = 720
✔ Check: First digit: 10 choices, second: 9 left, third: 8 left → 10×9×8=720
Answer for #2: 720
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Problem 3: How many ways can 3 people win 1st, 2nd, and 3rd place?
This implies there are more than 3 people total? Wait — actually, the problem doesn’t specify how many people are competing. But looking back at the example above: “There are 8 people running a race...” — so maybe this is continuing that context?
Wait — Problem 3 stands alone: “How many ways can 3 people win 1st, 2nd, and 3rd place?”
That phrasing is ambiguous. Does it mean:
A) There are only 3 people, and they are being assigned to 1st, 2nd, 3rd? → Then it’s 3! = 6
OR
B) There are more people (like 8 from previous example), and we’re choosing 3 winners out of them?
Looking at the worksheet structure — Problem 3 comes after Problem 2, and before Problem 4 (“8 students sit in 4 seats”). Also, the example above the table used 8 people for 1st/2nd/3rd.
But Problem 3 doesn’t mention a total number. Hmm.
Wait — perhaps it’s implying: “If there are 3 people, how many ways can they be ordered into 1st, 2nd, 3rd?” That would be 3!.
Alternatively, maybe it’s a trick — if only 3 people are running, then yes, 3! = 6.
But let’s look at the pattern:
- Example in table: 8 people → 1st, 2nd, 3rd → uses permutation without repetition: ₈P₃
- Problem 4: 8 students in 4 seats → ₈P₄
So likely, Problem 3 is meant to be similar — but it doesn’t say how many total people.
Wait — re-read: “How many ways can 3 people win 1st, 2nd, and 3rd place?”
This probably means: There are 3 distinct people, and we assign them to 3 distinct places → order matters → 3! = 6
Because if it were selecting 3 winners from a larger group, it would say “from X people”.
Also, in the absence of a specified total, the most logical interpretation is that we have exactly 3 people to assign to 3 positions.
So:
n = 3, r = 3, no repetition → 3! = 6
✔ Answer for #3: 6
*(Note: If the problem intended a larger pool, it should have specified. Given the wording, 6 is correct.)*
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Problem 4: How many ways can 8 students sit in 4 seats?
This is classic permutation without repetition.
- n = 8 students
- r = 4 seats
- Order matters (who sits where)
- No repetition (one student per seat)
Formula: n! / (n - r)! = ₈P₄
Calculate:
₈P₄ = 8 × 7 × 6 × 5 = ?
8 × 7 = 56
56 × 6 = 336
336 × 5 = 1680
✔ Answer for #4: 1680
---
Now, let’s compile final answers.
Final Answer:
1. 3125
2. 720
3. 6
4. 1680
Parent Tip: Review the logic above to help your child master the concept of permutation combination worksheet.