Questions: Permutations

Question 1

A club of 10 members needs to elect a president, vice president, and treasurer — three distinct roles. How many different outcomes are possible?

• A. 10^3 = 1,000, because there are 10 choices for each of the three roles
• B. 10! = 3,628,800, because all members must be assigned a position
• C. P(10,3) = 720, because the same member cannot hold two roles and the roles are distinct
• D. C(10,3) = 120, because we are selecting 3 people from 10

Answer: (C)

This is a permutation because the three roles are distinct — 'Alice-president, Bob-VP' is a different outcome from 'Bob-president, Alice-VP.' We fill 3 ordered slots from 10 candidates without replacement: 10 × 9 × 8 = P(10,3) = 720. Option A (10^3 = 1000) would be correct if the same person could hold multiple roles (repetition allowed). Option D (combinations) would be correct only if the roles were identical — which they are not.

Question 2

A student creates a 3-digit PIN where digits can repeat (e.g., 007 is valid). Another student counts the number of ways to arrange 3 different digits from 0–9 onto a sequence of numbered slots. Which student is solving a permutation problem?

• A. The first student, because PINs require order to matter
• B. The second student, because permutations require no repetition
• C. Both students, because both problems involve ordered arrangements
• D. Neither student, because both problems involve digits, not objects

Answer: (B)

The second student is solving a standard permutation problem: P(10,3) = 720 ordered arrangements of 3 distinct digits. The first student's PIN problem allows repetition, giving 10 × 10 × 10 = 10^3 = 1,000 — this is not a permutation. Order matters in both problems, but permutations specifically assume no repetition. The common misconception is that P(n,r) = n^r; in fact, n^r counts ordered arrangements *with* replacement, while P(n,r) = n!/(n-r)! counts them *without*.

Question 3

P(n,r) = n!/(n–r)! counts the number of ordered arrangements of r objects chosen from n distinct objects without replacement.

• T. True
• F. False

Answer: True

This is the definition of a permutation. Filling r ordered slots from n objects without replacement gives n choices for slot 1, n–1 for slot 2, ..., down to n–r+1 for slot r. The product n × (n–1) × ··· × (n–r+1) equals n!/(n–r)! because dividing n! by (n–r)! cancels the unwanted factorial terms.

Question 4

The number of ways to arrange most 5 books on a shelf equals P(5,3), because you are placing 5 objects in 3 possible positions.

• T. True
• F. False

Answer: False

When arranging all n objects, the correct formula is P(n,n) = n!/0! = n!. Arranging 5 books on a shelf gives 5! = 120, not P(5,3) = 60. P(5,3) would count ordered arrangements of only 3 books chosen from 5 — that is, filling 3 labeled slots while leaving 2 books aside.

Question 5

Why does P(n,r) = n!/(n–r)! rather than n^r? What scenario would n^r correctly count, and what makes it different from a permutation?

Model answer: P(n,r) = n!/(n–r)! counts ordered arrangements without replacement: each object can appear at most once, so choices shrink at each slot (n, then n–1, then n–2…). n^r counts ordered arrangements with replacement: each slot is filled independently from all n options, so the number of choices stays constant at n per slot. The difference is whether the same object can appear more than once. For example, arranging 3 different winners from 10 runners is P(10,3) = 720; generating a 3-digit code where digits can repeat is 10^3 = 1,000.

The formula P(n,r) = n!/(n–r)! is derived from the counting principle applied to shrinking choices. n^r treats each position as an independent choice, which is correct when repetition is allowed — license plates, PINs, passwords. Permutations model scenarios like race placements or officer elections where one object cannot occupy two positions simultaneously.