Permutations and Combinations · Lesson 2 of 6
Permutations of Distinct Objects
“Derive permutation counts from ordered positions and use factorials to write them efficiently.”
• Explain why exchanging positions creates a different permutation. • Derive the product for arranging r distinct objects chosen from n. • Use factorial notation and simplify factorial expressions. • Explain the special cases of arranging all objects and arranging none. • Distinguish permutations without repetition from ordered choices with repetition.
Order is part of the outcome
A chairperson and a vice-chairperson have different roles. Choosing A as chairperson and B as vice-chairperson produces a different result from choosing B as chairperson and A as vice-chairperson. An ordered letter string behaves in the same way: changing positions changes the result.
A permutation is an arrangement of some or all available objects in a definite order. For distinct objects without repetition, each chosen object occupies one position only.
The notation ⁿPᵣ means the number of permutations of n different objects taken r at a time, without repetition. Here n is the number available, while r is the number of positions filled. Both are whole numbers and 0 ≤ r ≤ n. Read the notation as “n P r”; the superscript is a count of objects, not an exponent.
Deriving the position product
Start with r empty positions and n different objects. The first position has n choices. Whichever object is chosen, the next position has n − 1 choices. Continue removing one used object at each step. At position r, exactly r − 1 objects have already been used.
The last position therefore has n − (r − 1) = n − r + 1 choices. There are r factors in the product. This is an application of the multiplication principle, rather than an unrelated rule to memorise.
Problem
How many three-letter strings can be made from the letters of NUMBER without repeating a letter?
- 1.NUMBER contains 6 different letters. The strings have 3 ordered positions.
- 2.The position counts are 6, 5 and 4. Thus ⁶P₃=6×5×4=120.
- 3.The result counts strings with or without meaning. NUM and MUN are different because their orders differ.
Factorials make the product compact
Long products occur frequently in arrangement problems. Factorial notation gives a short name to the product of all positive whole numbers up to a chosen number. It is useful both for expressing a count and for cancelling common factors.
For a positive whole number n, n! is the product 1 × 2 × 3 × ⋯ × n. The symbol is read “n factorial”. We also define 0! = 1.
A factorial does not mean repeated multiplication of one number. For example, 4! = 4 × 3 × 2 × 1 = 24, whereas 4⁴ = 256. Factorials also do not distribute over addition or subtraction: calculate each factorial or factor the expression legitimately.
Problem
Evaluate 5!, 7! and 7! − 5!.
- 1.5! = 1 × 2 × 3 × 4 × 5 = 120.
- 2.7! = 7 × 6 × 5! = 42 × 120 = 5040.
- 3.Subtract the evaluated quantities: 7! − 5! = 5040 − 120 = 4920. It is not (7 − 5)! = 2.
Problem
Evaluate 7!/5! and 12!/(10!2!).
- 1.Expand only as far as the denominator: 7! = 7 × 6 × 5!. Dividing by 5! leaves 7 × 6 = 42.
- 2.Similarly, 12! = 12 × 11 × 10!, while 2! = 2. Thus 12!/(10!2!) = (12 × 11)/2 = 66.
- 3.Cancellation removes equal multiplicative factors. It does not permit cancelling the factorial symbols themselves.
Problem
Evaluate n!/[r!(n − r)!] for n = 5 and r = 2.
- 1.Substitute both values, including their difference: 5!/[2!(5 − 2)!] = 5!/(2!3!).
- 2.Expand 5! as 5 × 4 × 3!, cancel 3!, then divide by 2! = 2. The value is (5 × 4)/2 = 10.
- 3.This expression will later count unordered selections. Here the goal is careful substitution and factorial arithmetic.
Problem
If 1/8! + 1/9! = x/10!, find x.
- 1.Multiply every term by 10!: x = 10!/8! + 10!/9!.
- 2.The first quotient is 10 × 9 = 90. The second quotient is 10.
- 3.Therefore x = 90 + 10 = 100. Multiplying by a common denominator avoids adding unlike denominators.
The factorial form of the permutation count
The descending product stops at n − r + 1. The missing factors below it form (n − r)!. Multiplying our product by that factorial completes n!, so dividing n! by the missing factorial leaves precisely the r factors we need.
For 1 ≤ r < n, write n! = n(n − 1)⋯(n − r + 1)(n − r)!. Dividing both sides by (n − r)! proves the formula. If r = n, every object is arranged and the denominator is 0! = 1, giving n!. If r = 0, there is one empty arrangement: doing nothing. The formula gives n!/n! = 1, which matches that interpretation.
Problem
From 12 people, choose a chairperson and a vice-chairperson. No person may hold both offices.
- 1.The two offices are ordered roles, so we need ¹²P₂, rather than an unordered pair count.
- 2.Choose the chairperson in 12 ways and the vice-chairperson in 11 ways.
- 3.The number is 12 × 11 = 132 = 12!/10!.
When repetition is allowed
If a symbol can be chosen again, filling a position does not reduce the number of symbols available. With n symbols and r positions, every position has n choices. Multiply n by itself r times.
For NUMBER, three-letter strings with repetition number 6³ = 216. Without repetition the count is 120. For ROSE, all four letters without repetition give 4! = 24 arrangements, while four positions with repetition give 4⁴ = 256. Always state which model the problem uses before substituting numbers.
The formula n!/(n − r)! is valid here only for whole numbers with 0 ≤ r ≤ n. A task cannot choose more distinct objects than are available. A numerical expression outside that domain does not represent a valid permutation count.
Quiz
A chairperson and secretary are chosen from 7 people, with different people in the roles. How many outcomes are possible?
What is 0!?
What is 9!/7!?
How many four-position strings use six symbols when repetition is allowed?
Why is ⁿP₀=1?
Practice Problems
- Evaluate 8! and 4! − 3! without assuming factorials distribute over subtraction.
- Evaluate 8!/(6!2!) by cancelling products.
- Solve 1/6! + 1/7! = x/8!.
- Evaluate n!/(n − r)! for (n, r) = (6, 2) and (9, 5).
- Count two different offices filled from eight people. Explain why the result differs from choosing an ordinary pair.
- Compare three-letter codes from six letters with and without repetition.
- Explain in words why arranging all n distinct objects gives n! and arranging none gives one outcome.
Key Takeaways
• A permutation records the order or role of each chosen object. • The product n(n − 1)⋯(n − r + 1) follows from filling positions. • Factorial cancellation shortens the same product to n!/(n − r)!. • The empty arrangement explains the value 1 when r = 0. • With repetition allowed, n symbols in r positions give nʳ outcomes.