A permutation is a selection that also keeps track of the order in which the chosen items are lined up.
Pick the same three people, and if who stands first, second or third changes, it counts as a different case.
So the first place has n choices, the next has one fewer, n − 1, and multiplying all the way to place r gives ₙPᵣ.
Move the n and r sliders above and the choices left at each place and their product appear at once; the round-table slider counts circular permutations, where rotations count as the same.
A Selection with Order
Choosing the same people still gives more cases once the places are ranked. A permutation counts those place orders too, and ₙPᵣ is the product of how many items are left at each place when r places are filled from the front out of n items. The first place has n, the next n − 1, and so on down to place r, and that product equals n! ÷ (n − r)!. Lining everyone up is the case r = n, which gives n!, and when some items are identical you must divide by the swaps that cannot be told apart. At a round table everyone moving one seat over gives the same seating, so the n rotations count as one and the value is (n − 1)!. Setting 0! = 1 is the agreement that keeps the formula correct when r = n.
🎯 What a permutation is
①The number of ways to choose r of n distinct items and line them up in a chosen order is written ₙPᵣ
②As when choosing first, second and third place, who stands in which place changes the result
③Even the same three people form a different case if they stand in a different order
④Combinations, where order does not matter, come in the next chapter
Counting Place by Place
5
3
seegongsik.com
The choices left at each place when r of n items fill the places from the 1st
At the top lie n numbered balls, and only the first r are colored. Below them runs a row of places from the 1st to the rth, each showing how many items are still left to choose for that place: n, n − 1, and so on. Dashed lines join the balls above to the places below, and the bottom line writes ₙPᵣ out as that product. n moves from 2 to 8 and r from 1 to n, starting at n = 5, r = 3. In the round-table figure, as many colored balls as there are people sit in a ring, and ball 1, whose seat is fixed, carries the mark "fix". Under the figure the line count n! is divided by the n rotations to give (n − 1)!; the table seats 3 to 7 people and starts at 4.
💡 Multiplication principle
①The first place can take any of n items
②One item is already used, so n − 1 are left for the second place
③Each place loses one more, so n − r + 1 are left for place r
④Multiplying them all gives n × (n − 1) × … × (n − r + 1)
Number of permutations
nPr = n!(n-r)!
The number of ways to choose r of n items and line them up in order
Special Permutations
Counting n people at a round table as n! for a line counts each seating n times, once for every shift by one seat. To merge the seatings that rotation makes identical, fix one person and line up only the other n − 1, which gives (n − 1)!. Using combinations, which erase order, for a problem about forming a line drops the lines that differ only in order, so the answer comes out too small. Two people who must stand together are first tied into one block and lined up, and then the 2! ways they can swap inside the block are multiplied in. Leaving out the denominator in an arrangement with identical items counts swaps that cannot be told apart as different cases.
Lining up all n
nPn = n!
The number of ways to line up all n items in a row
Permutations with identical items
n!p! q! r!
When p, q and r of the n items are identical. Swapping identical items changes nothing, so divide by those counts
Circular Permutations
4
seegongsik.com
n people seated at a round table, and dividing the line count by the number of rotations
🔄 The heart of circular permutations
①Around a circle, everyone moving one seat over gives the same seating
②So fix one person's seat and line up the rest
③The number of circular permutations of n people is (n − 1)!
④It can also be seen as the n! lines overlapping n at a time through rotation, giving n! ÷ n = (n − 1)!
Number of circular permutations
(n−1)!
The number of ways to arrange n items in a circle
Worked Examples
Example 1
Find ₅P₂, the number of ways to arrange 2 of 5 distinct items in a row.
1
Multiply 5 choices for the first place by 4 for the second.
₅P₂ = 5 × 4
2
Compute the product.
= 20
▸ 20
₅P₂ = 20 is 2 times ₅C₂ = 10, because the two chosen items can be lined up in 2! = 2 orders.
Example 2
In how many ways can 5 people stand in a row so that 2 particular people stand next to each other?
1
Tie the 2 people into one block; lining up the 4 units gives 4! ways.
block arrangement 4! × internal order 2!
2
Multiply by the 2! ways the two can swap inside the block.
= 24 × 2 = 48
▸ 48
For a "next to each other" condition, line up the block first, then multiply by the order inside it.
Summary
Number of permutations
nPr = n!(n-r)! = n(n-1)(n-2)···(n-r+1)
The number of selections in which order changes the result
Provincial mock exam type
How many three-digit numbers can be made by choosing 3 different digits from the 5 digits 1, 2, 3, 4, 5?
①10
②20
③30
④60
⑤120
▸ ④ 60
1
Three of 5 different digits are chosen and placed in the hundreds, tens and ones places, so the count is ₅P₃.
₅P₃ = 5 × 4 × 3
2
Compute the product.
= 60
🎯 Exam Points
①If order matters it is a permutation; if not, a combination
②ₙPₙ = n! counts lining up everything
③The number of circular permutations is (n − 1)!
④For identical items, divide the factorial of the total by the factorials of the identical groups