Permutation Formula Quiz: Basic Permutation Formula

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Quizzes Created: 7682 | Total Attempts: 9,547,133
| Questions: 20 | Updated: Dec 17, 2025
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1) Compute P(5, 2).

Explanation

P(5,2)=5!/3!=5×4=20.

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About This Quiz
Permutation Formula Quiz: Basic Permutation Formula - Quiz

How can you count arrangements when the order of items matters? In this quiz, you’ll explore the permutation formula and learn how to apply it to situations involving ordered selections. You’ll practice identifying when permutations are appropriate, calculating results step by step, and interpreting what each term represents. Through clea... see moreexamples, you’ll develop a stronger sense of how permutations model arrangements and how to approach counting problems that rely on ordering.
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2)
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2) P(6, 3) equals 6×5×4 = 120.

Explanation

P(6,3)=6!/3!=6×5×4=120, so the statement is true.

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3) Compute P(7, 1). The value is ____.

Explanation

P(7,1)=7!/6!=7.

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4) Compute P(8, 0).

Explanation

P(n,0)=n!/n!=1 for any n≥0.

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5) Select all expressions equal to P(9, 2).

Explanation

P(9,2)=9!/7!=9×8. 9C2 counts unordered pairs; 9^2 is not a permutation count without restriction.

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6) Compute P(6, 6).

Explanation

P(6,6)=6!/0!=6!=720.

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7) If r>n in permutations without repetition, then there are 0 valid arrangements.

Explanation

You cannot choose and order more distinct items than available, so the count is 0.

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8) Compute P(10, 2). The value is ____.

Explanation

P(10,2)=10!/8!=10×9=90.

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9) How many 3-letter arrangements can be made from 5 distinct letters?

Explanation

Order matters without repetition: P(5,3)=5×4×3=60.

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10) Which are equal to P(7, 3)? Select all that apply.

Explanation

P(7,3)=7!/4!=7×6×5. Also combinations×3! gives the same: 7C3×3!=7P3. 7^3 is not correct here.

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11) For any n≥1, P(n, 1)=n.

Explanation

P(n,1)=n!/ (n−1)! = n.

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12) If P(5, r)=20, then r equals ____.

Explanation

Compute P(5,2)=5×4=20, while P(5,1)=5 and P(5,3)=60. So r=2.

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13) Compute P(9, 4).

Explanation

P(9,4)=9!/5!=9×8×7×6=3024.

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14) Select all expressions equal to 5!.

Explanation

5!=120. Also P(5,5)=5!, and 5×P(4,4)=5×4!=5!. P(6,5)=6!/1!=720, P(10,0)=1.

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15) From 12 runners, how many ways to award gold, silver, and bronze (distinct places)?

Explanation

Order matters: P(12,3)=12×11×10=1320.

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16) P(n, 0)=0 for all n≥0.

Explanation

P(n,0)=n!/n!=1, so the statement is false.

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17) Compute P(4, 3). The value is ____.

Explanation

P(4,3)=4!/1!=4×3×2=24.

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18) If P(n, 2)=90, what is n?

Explanation

n(n−1)=90 ⇒ n^2−n−90=0 ⇒ (n−10)(n+9)=0, so n=10 (positive solution).

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19) Select all expressions equal to the number of ways to pick and order 2 from 5.

Explanation

All listed except 2^5 equal P(5,2)=5×4=20.

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20) Choosing a class president and vice-president from 7 students (distinct roles) equals:

Explanation

Order matters because roles differ: P(7,2)=7×6=42.

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Compute P(5, 2).
P(6, 3) equals 6×5×4 = 120.
Compute P(7, 1). The value is ____.
Compute P(8, 0).
Select all expressions equal to P(9, 2).
Compute P(6, 6).
If r>n in permutations without repetition, then there are 0 valid...
Compute P(10, 2). The value is ____.
How many 3-letter arrangements can be made from 5 distinct letters?
Which are equal to P(7, 3)? Select all that apply.
For any n≥1, P(n, 1)=n.
If P(5, r)=20, then r equals ____.
Compute P(9, 4).
Select all expressions equal to 5!.
From 12 runners, how many ways to award gold, silver, and bronze...
P(n, 0)=0 for all n≥0.
Compute P(4, 3). The value is ____.
If P(n, 2)=90, what is n?
Select all expressions equal to the number of ways to pick and order 2...
Choosing a class president and vice-president from 7 students...
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