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high school 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 and the choices shrink at each spot, and nPr appears; use the round-table slider and circular permutations, where rotations count as the same, appear.

Order Matters

Picking the same people still yields more cases once you assign seats. A permutation is the count that keeps those seat orders, and the symbol nPr is the product of the leftovers as you fill r seats from n items, front to back. The first seat has n choices, the next n − 1, and so on through seat r; that product is the same as n! / (n − r)!. At a round table, turning the whole seating looks the same, so n rotations count as one and the value is (n − 1)!. Lining everyone up is the case r = n, which is n!. If some items match, you divide the full factorial by the factorials of those repeats. 0! = 1 because these formulas treat an empty product as 1.

🎯 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
④Combinations (order does not matter) are in the next chapter

Visualize nPr

5
3

The upper figure lays out n balls with numbers; only the first r are filled with color. Below, r slots run from the 1st onward, and each slot shows the leftover count n, n − 1, and so on. Dashed lines join the balls to the slots, and the bottom line writes nPr as that product. n runs from 2 to 8 and r from 1 to n. The starting values are n = 5 and r = 3. On the round-table figure, as many colored balls as the chosen count sit on a circle, with the word for a fixed seat beside the top place. The formula under the circle shows the value of (n − 1)!. The table count runs from 3 to 7 and starts at 4.

💡 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

If you count n people around a table as if they were a line of n!, you double-count seatings that are just one rotation of each other. To treat those rotations as one seating you fix one person and arrange the rest, and that value is (n − 1)!. On a lining-up problem, a symbol that drops order throws away swapped front-and-back lines and leaves the count too small. Two people who must sit together are first treated as one block, then the block’s own front and back go into the product. In a listing with repeated items, skipping the denominator counts indistinguishable seats as different cases.

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. Formula box only — the sketches are nPr and circular

Circular Permutations

4
🔄 Why (n−1)!
①Rotation gives the same arrangement
②Fix one person, arrange the rest
③Circular permutation count = (n−1)!
④n rotations coincide, so n!/n = (n−1)!
⑤0! = 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
②P(n,n) = n!
③Circular = (n−1)!
④Repeated items: n! / (p!q!r!)
⑤Remember 0! = 1
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Applications of Definite Integrals
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Combination
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