Permutation Formula Quiz: Basic Permutation Formula

The answer is True. P(n,1) = n! divided by (n-1)! = n. Selecting and ordering 1 item from n gives exactly n outcomes — one for each possible item chosen. This is consistent with the general formula and holds for all positive integers n.

The two roles are distinct, so order matters — being president is different from being vice-president. P(7,2) = 7 times 6 = 42. Option A uses combinations, which ignores the distinction between the two roles and undercounts by half. Option C gives 49 = 7 squared, allowing the same person for both roles. Option D gives P(7,3) which selects 3 people for only 2 roles.

P(5,2) = 20. 5 times 4 = 20, confirming A. 5C2 times 2! = 10 times 2 = 20, confirming B since multiplying combinations by r! recovers the permutation. 5! divided by 3! = 120 divided by 6 = 20, confirming C. 5P2 is the standard notation for P(5,2) = 20, confirming D.

P(n,2) = n times (n-1) = 90. Solving: n squared minus n minus 90 = 0, giving (n-10)(n+9) = 0, so n = 10. Check: 10 times 9 = 90. Option A: 8 times 7 = 56, not 90. Option B: 9 times 8 = 72, not 90. Option C: 11 times 10 = 110, not 90. Only n=10 satisfies the equation.

P(4,3) = 4! divided by (4-3)! = 4! divided by 1! = 24 divided by 1 = 24. Equivalently 4 times 3 times 2 = 24. Option A gives 12 = 4 times 3, missing the third factor. Option B gives 18, which has no direct relationship to the formula. Option D gives 36 = 6 squared, also unrelated.

The answer is False. P(n,0) = n! divided by n! = 1 for all valid n. There is exactly one way to arrange zero items — the empty arrangement. The statement claims the answer is 0, which is incorrect. Confusing P(n,0) with having no arrangements is a common error.

The three medals are distinct, so order matters. P(12,3) = 12 times 11 times 10 = 1320. Option A gives 660 = P(12,3) divided by 2, which incorrectly halves the count. Option B gives 1188, which has no valid derivation. Option D gives P(12,4) = 11880 divided by something — 1716 = C(13,4) which is unrelated. Only 1320 correctly applies the permutation formula.

5! = 120, confirming A. P(5,5) = 5! divided by 0! = 120, confirming B. 5 times P(4,4) = 5 times 4! = 5 times 24 = 120, confirming C. Option D: P(6,5) = 6! divided by 1! = 720, which equals 6! not 5!, so it does not equal 5!.

P(9,4) = 9 times 8 times 7 times 6 = 3024. Option B gives 362880 = 9!, which would be P(9,9). Option C gives 1512, which is half of 3024 with no valid derivation. Option D gives 504 = 9 times 8 times 7, which is P(9,3) not P(9,4). Only 3024 correctly multiplies four descending factors from 9.

P(5,1) = 5, P(5,2) = 5 times 4 = 20, P(5,3) = 5 times 4 times 3 = 60. Only r=2 gives 20. Option A gives P(5,1)=5, not 20. Option C gives P(5,3)=60. Option D gives P(5,4)=120. Only r=2 satisfies P(5,r)=20.

P(5,2) = 5! divided by (5-2)! = 5! divided by 3! = 5 times 4 = 20. Option A gives 15, option B gives 10, option D gives 30, none of which correctly apply the permutation formula. The first selection has 5 choices and the second has 4, giving 5 times 4 = 20 ordered arrangements.

P(7,3) = 7 times 6 times 5 = 210, confirming A. 7! divided by 4! = 210, confirming B. 7P3 is the standard notation for P(7,3), confirming C. 7C3 times 3! = 35 times 6 = 210, confirming D since combinations times the number of orderings of r items always recovers the permutation count.

Order matters and repetition is not allowed, so use P(5,3) = 5 times 4 times 3 = 60. The first position has 5 choices, second has 4, third has 3. Option A gives 50, option C gives 80, option D gives 100, none of which correctly multiply the descending factors.

P(10,2) = 10! divided by 8! = 10 times 9 = 90. Option A gives 45 = 10 times 9 divided by 2, which is the combination C(10,2) not the permutation. Option B gives 80, option D gives 100, neither of which correctly apply the descending product formula.

The answer is True. You cannot select and order more distinct items than are available. If r exceeds n, the formula requires computing a factorial of a negative number in the denominator, which is undefined. In practice this means no valid arrangements exist, and the count is defined as 0.

P(6,6) = 6! divided by (6-6)! = 6! divided by 0! = 720 divided by 1 = 720. This counts all possible orderings of 6 distinct items, which equals 6! = 720. Option A gives 36 = 6 squared. Option B gives 120 = 5!. Option C gives 600, which has no direct relationship to 6!.

P(9,2) = 9 times 8 = 72, confirming A. 9! divided by 7! = 9 times 8 = 72, confirming B. 9P2 is simply another notation for P(9,2), confirming C. Option D gives 9C2 = 36, which counts unordered pairs and is half of P(9,2) — combinations do not account for order and give a different result.

P(n,0) = n! divided by n! = 1 for any n greater than or equal to 0. There is exactly one way to arrange zero items — the empty arrangement. Option A gives 0, which would incorrectly suggest no arrangements exist. Options C and D misapply the formula entirely.

P(7,1) = 7! divided by (7-1)! = 7! divided by 6! = 7. Choosing and ordering 1 item from 7 gives exactly 7 possibilities — one for each item. Option A gives 1, option B gives 6, option D gives 42, none of which correctly apply the formula with r=1.

The answer is True. P(6,3) = 6! divided by (6-3)! = 6! divided by 3! = 6 times 5 times 4 = 120. The formula produces a descending product starting from n and taking r factors. With n=6 and r=3, the product is 6 times 5 times 4 = 120.