Mathematics from zero
Permutations
Three runners finish a race. Gold, silver, bronze — but in what order? Any of the 3
could take gold; then either of the remaining 2 takes silver; the last takes bronze.
3 × 2 × 1 = 6 possible podiums. That count of orderings is a permutation.
After this lesson you can say what a permutation is, count the arrangements of a set of items, use the factorial, and count arrangements when you only place some of the items.
A permutation is an arrangement in which order matters. Lining up 3 runners on a podium, ordering books on a shelf, choosing a first and second prize — these all care about which position each item takes. A permutation is one such ordered arrangement, and the question is how many there are.
Count permutations with the counting principle — but the options shrink. The first
position has all n items to choose from. Once one is placed, the second position has
only n − 1 left, the third n − 2, and so on. By the counting principle you multiply
these shrinking counts: for 3 runners, 3 × 2 × 1.
Multiplying a whole number down to 1 is called a factorial. The product
n × (n − 1) × ... × 2 × 1 is the factorial of n, written n!. So 3! = 6 and
4! = 24. The number of ways to arrange all n items in order is exactly n! —
factorials and full arrangements are the same count.
To arrange only some of the items, stop multiplying early. If you place just r
items out of n, you only need the first r shrinking factors. Choosing a 1st and
2nd prize from 5 people fills 2 positions: 5 × 4 = 20. You take as many factors as
positions, starting from n and counting down.
In how many orders can 4 different books be arranged on a shelf?
All 4 books are placed, and order matters — this is a permutation of all 4.
The leftmost spot has 4 books to choose from. The next spot has 3 left, then 2, then
the last book fills the final spot — 1 way. Multiply the shrinking counts:
4 × 3 × 2 × 1.
That product is 4!, and 4! = 24. There are 24 orders. Notice how fast it climbs:
3! = 6, 4! = 24, 5! = 120 — factorials grow even faster than exponents.
Why this works
Why do the choices shrink by one each time? Because an item, once placed, is used up — it cannot appear in a second position. The first position drew from the full set; the second draws from whatever is left. Each placement removes one item, so the pool of choices drops by exactly one at every step. That steady shrink is what produces the factorial.
Common mistake
A common mistake is using the full factorial when you are only placing some of the
items. Arranging 3 of 5 people is 5 × 4 × 3 = 60, not 5! = 120. Take only as many
shrinking factors as there are positions to fill. 5! would arrange all five, which is
a different question.
Work out 3! (that is 3 × 2 × 1). Type the value.
Work out 4!. Type the value.
Arrange 3 of 5 people in order: 5 × 4 × 3. Type the count.
Work out 5!. Type the value.
In how many orders can 2 items be arranged? Type the count.
Why is the number of ways to arrange 4 items equal to 4 × 3 × 2 × 1?
A permutation is an arrangement where order matters. Count permutations with the
counting principle, but note the choices shrink by one at each position as items are
used up. Multiplying a whole number down to 1 is the factorial, n!, and n! is the
number of ways to arrange all n items. To arrange only r of them, take just the
first r shrinking factors.