This lesson gives each rule, the reason it works and seven worked examples. It also lists the traps that cost marks.
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Q1. Evaluate 4P3.
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Answer: D
We need 4P3, permutations of 4 items taken 3 at a time. We’re arranging 3 items out of 4, where order matters, so we use permutations. nPr = n! / (n-r)!, so 4P3 = 4! / 1!. This is the product of 3 numbers counting down from 4: 4 x 3 x 2 = 24. So 4P3 = 24.
Related topics: Permutations and Combinations: Formulas, Shortcuts and Solved Examples
Q2. Evaluate 9P4.
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Answer: C
We need 9P4, permutations of 9 items taken 4 at a time. We’re arranging 4 items out of 9, where order matters, so we use permutations. nPr = n! / (n-r)!, so 9P4 = 9! / 5!. This is the product of 4 numbers counting down from 9: 9 x 8 x 7 x 6 = 3024. So 9P4 = 3024.
Related topics: Permutations and Combinations: Formulas, Shortcuts and Solved Examples
Q3. Evaluate 8P4.
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Answer: B
We need 8P4, permutations of 8 items taken 4 at a time. We’re arranging 4 items out of 8, where order matters, so we use permutations. nPr = n! / (n-r)!, so 8P4 = 8! / 4!. This is the product of 4 numbers counting down from 8: 8 x 7 x 6 x 5 = 1680. So 8P4 = 1680.
Related topics: Permutations and Combinations: Formulas, Shortcuts and Solved Examples
Q4. Evaluate 11P3.
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Answer: D
We need 11P3, permutations of 11 items taken 3 at a time. We’re arranging 3 items out of 11, where order matters, so we use permutations. nPr = n! / (n-r)!, so 11P3 = 11! / 8!. This is the product of 3 numbers counting down from 11: 11 x 10 x 9 = 990. So 11P3 = 990.
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Q5. Evaluate 7P4.
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Answer: A
We need 7P4, permutations of 7 items taken 4 at a time. We’re arranging 4 items out of 7, where order matters, so we use permutations. nPr = n! / (n-r)!, so 7P4 = 7! / 3!. This is the product of 4 numbers counting down from 7: 7 x 6 x 5 x 4 = 840. So 7P4 = 840.
Related topics: Permutations and Combinations: Formulas, Shortcuts and Solved Examples
Q6. Evaluate 13P3.
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Answer: D
We need 13P3, permutations of 13 items taken 3 at a time. We’re arranging 3 items out of 13, where order matters, so we use permutations. nPr = n! / (n-r)!, so 13P3 = 13! / 10!. This is the product of 3 numbers counting down from 13: 13 x 12 x 11 = 1716. So 13P3 = 1716.
Related topics: Permutations and Combinations: Formulas, Shortcuts and Solved Examples
Q7. Evaluate 9!
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Answer: C
We need 9 factorial, written 9!. 9! means multiply every whole number from 9 down to 1. 9! = 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 362880. So the answer is 362880.
Related topics: Permutations and Combinations: Formulas, Shortcuts and Solved Examples
Q8. Evaluate 7!
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Answer: A
We need 7 factorial, written 7!. 7! means multiply every whole number from 7 down to 1. 7! = 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5040. So the answer is 5040.
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Q9. Evaluate 10!
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Answer: B
We need 10 factorial, written 10!. 10! means multiply every whole number from 10 down to 1. 10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 3628800. So the answer is 3628800.
Related topics: Permutations and Combinations: Formulas, Shortcuts and Solved Examples
Q10. Evaluate 4!
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Answer: D
We need 4 factorial, written 4!. 4! means multiply every whole number from 4 down to 1. 4! = 4 x 3 x 2 x 1 = 24. So the answer is 24.
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Q11. Evaluate 5!
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Answer: A
We need 5 factorial, written 5!. 5! means multiply every whole number from 5 down to 1. 5! = 5 x 4 x 3 x 2 x 1 = 120. So the answer is 120.
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Q12. Evaluate 8!
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Answer: C
We need 8 factorial, written 8!. 8! means multiply every whole number from 8 down to 1. 8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40320. So the answer is 40320.
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Q13. Evaluate 6!
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Answer: B
We need 6 factorial, written 6!. 6! means multiply every whole number from 6 down to 1. 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720. So the answer is 720.
Related topics: Permutations and Combinations: Formulas, Shortcuts and Solved Examples