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Grade 11-12 (age 16-18)

Permutation

Permutation

A permutation is a selection where the order matters. Even if you pick the same three people, placing them in first, second, or third counts as a different case. So the first spot has n choices, the next one fewer at (n-1), and so on; multiply r of them and you get nPr. Drag the top sliders for n and r to watch the choices shrink at each spot and read off nPr, and use the round-table slider to explore circular permutations, where rotations count as the same.

Order Matters
🎯 What is a Permutation?
①Pick r out of n distinct items and arrange in order
②1st/2nd/3rd places — who goes where matters
③Three of the same people but different orders count as different arrangements
Visualize nPr
5
3
💡 Multiplication Rule
①1st slot: n choices
②2nd slot: (n−1) choices
③Choices decrease toward the r-th slot
④Result: n × (n−1) × … × (n−r+1)
Permutation Formula
nPr = n!(n-r)!
Pick r from n with ordering
Special Permutations
Arrange All n
nPn = n!
Number of orderings of n items
With Repeated Items
n!p! q! r!
When there are p, q, r repeats
Circular Permutations
4
🔄 Why (n−1)!
①Rotation gives the same arrangement
②Fix one person, arrange the rest
③Circular permutation count = (n−1)!
Circular Permutation
(n−1)!
Arrange n items in a circle
Work It Out
Example 1
Find ₅P₂, the number of ways to arrange 2 of 5 distinct items in a row.
1
A permutation multiplies, starting from the largest number, as many factors as items chosen.
₅P₂ = 5 × 4
2
Compute.
= 20
20
A permutation counts arrangements where order matters.
Example 2
In how many ways can 5 people stand in a row so that 2 specific people are adjacent?
1
Treat the adjacent pair as one block, giving 4 items to arrange (4!).
block arrangement 4! × internal order 2!
2
Compute.
= 24 × 2 = 48
48
For an adjacency condition, bundle them, arrange, then multiply by the internal order.
Wrap-up
Core Formula
nPr = n!(n-r)! = n(n-1)(n-2)···(n-r+1)
Ordered selection count
Education-office assessment type
How many three-digit numbers can be formed by choosing 3 different digits from 1, 2, 3, 4, 5?
10
20
30
60
120
④ 60
1
Choosing and arranging 3 of 5 distinct digits gives ₅P₃.
₅P₃ = 5 × 4 × 3
2
Compute.
= 60
🎯 Exam Points
①P: order matters; C: order does not
②_nP_n = n!
③Circular = (n−1)!
④Repeated items: n! / (p!q!r!)
⑤Remember 0! = 1
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Combination
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