MATC44H3 Chap.8 Permutations, Combinations and the Selection Table
Permutations, Combinations and the Selection Table
Permutations and factorials
Any ordering of distinct objects in a row is a permutation of them; n objects have n! orderings, with 0! = 1. P(n, k), the number of ordered arrangements of k objects chosen from n, is n!/(n − k)!, a product of k falling factors starting at n, and it is 0 when k exceeds n.
The lecture proves the formula by filling k slots in turn.
Binomial coefficients
The binomial coefficient C(n, k), read n choose k, counts the subsets of size k inside a set of size n and equals n!/(k!(n − k)!). The slides derive it from P(n, k) = C(n, k) · k!, since an ordered selection is a subset followed by an ordering of its elements.
Forming a committee of four from nine people therefore gives C(9, 4) = 126, while assigning four named offices gives 9!/5! = 3,024.
The selection table
Selecting k objects from n is decided by two questions: does order matter, and is repetition allowed?
Ordered with repetition gives n to the power k; ordered without repetition gives P(n, k); unordered without repetition gives C(n, k); unordered with repetition gives C(n + k − 1, k), an entry the lecture flags as requiring a proof. Key words such as list, PIN, row and committee positions usually signal order; set, group, pile and poker hand usually signal that order does not matter. People are always treated as distinct.
Conditions such as at least or not both are handled by exact cases or by subtracting the forbidden selections.
What this chapter covers
- 01
Permutations and the factorial
- 02
Ordered selections P(n, k)
- 03
Binomial coefficients and subsets
- 04
The order and repetition table
- 05
Words that signal order
- 06
Committees with and without offices
- 07
At-least conditions and complements
Worked example · free
Four-digit PINs with conditions
- 2No repetition, order matters: 10 × 9 × 8 × 7 = 5,040 PINs.
- 1Increasing order: each set of four distinct digits gives exactly one increasing PIN.
- 1So the count is C(10, 4) = 210.
Key terms
- Permutation
- A permutation is one ordering of a collection of distinct objects.
- Factorial
- The factorial n! is the product n(n − 1)⋯2 · 1, with 0! defined as 1.
- Ordered Selection
- An ordered selection is a list of chosen objects in which the order of the entries matters.
- Complement Count
- A complement count finds the wanted selections by subtracting the unwanted ones from the total.
- Committee With Offices
- A committee with offices is a selection whose members hold named positions, so order effectively matters.
Permutations, Combinations and the Selection Table FAQ
How do I decide whether order matters in a counting question?
Ask whether two outcomes that contain the same objects in a different order are counted as different. Lists, rows, PINs, words and offices make them different. Sets, groups, committees without roles and hands of cards make them the same. When in doubt, test with a tiny case and list.
Why is C(n, k) equal to P(n, k) divided by k factorial?
Every ordered selection of k objects consists of a subset of size k together with one of its k! orderings. So P(n, k) = C(n, k) · k!, and dividing by k! removes the orderings. The relation is also a quick check on any answer that mixes the two.
When should I subtract instead of adding cases?
Subtract when the unwanted selections are fewer or simpler than the wanted ones, as with at least one or not both conditions. Count all selections, count the forbidden ones, and subtract. Add exact cases when the condition splits naturally into a few disjoint sizes.
Why does the selection table say people are always distinct?
People have names, so swapping two people produces a different outcome whenever their positions or roles differ. Questions about choosing people therefore never treat them as identical, unlike questions about identical balls or repeated letters, where only how many of each kind matters, not which one.
Exam move
Copy the selection table from memory and fill each cell with a small example you can list. Then solve one committee question both by exact cases and by subtraction, and check that the two answers agree. Finish with one PIN or licence plate question with and without repetition.
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