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ESO Y4A · Mathematics
combinatorics · Combinatorics

Factorial

In the previous section the same product appeared again and again: $n \cdot (n-1) \cdot (n-2) \cdots 1$. It comes up so often in combinatorics that it has its own name and its own symbol: the factorial. It is worth mastering, because it will go with you through everything that follows.

Before you start, make sure you know: Algebraic expressions (2nd ESO)
Before you start, make sure you know: First-degree equations (3rd ESO)

What is the factorial?

Table of factorial values from 0 to 7 illustrating its rapid growth 0!=1 1!=1 2!=2 · 1 = 2 3!=3 · 2 · 1 = 6 4! = 24 5! = 120 6! = 720 7! = 5 040 grows far faster than n

Notice how quickly it grows: going from $6!=720$ to $7!=5\,040$ you only multiply by $7$, but the jump is huge. That is why the numbers in combinatorics shoot up so fast.

The factorial is not arbitrary: it is exactly the number of permutations of $n$ distinct elements that you saw in the previous section, $P_n = n!$.

Practice★☆☆Not started
Calculate the exact value of the expression $6! - 4!$.

Calculate the exact value of the expression $6! - 4!$.

Solve

Simplifying fractions of factorials

Calculating $100!$ by hand would be absurd, but often there is no need: when the large factorial is divided by a smaller one, almost everything cancels.

Expansion and cancellation of common factors when simplifying a fraction with factorials 8! = 8 · 7 · 6! 8 · 7 · 6! 6! = 56 the 6! (in grey) cancels

The same trick works in reverse, to write a product as a quotient of factorials:

Practice★☆☆Not started
Simplify the following fraction of factorials without using a calculator: $\frac{8!}{6!}$.

Simplify the following fraction of factorials without using a calculator: $\frac{8!}{6!}$.

Solve
Practice★☆☆Not started
Express the numerical product $12 \cdot 11 \cdot 10$ using a division of factorials.

Express the numerical product $12 \cdot 11 \cdot 10$ using a division of factorials.

Solve

Factorials with letters

The same technique works when the number is unknown and is written as a letter. In combinatorics exercises it is common to have to simplify expressions such as $\dfrac{n!}{(n-2)!}$.

Algebraic breakdown of n! highlighting the factors n and n-1 and the cancellation of (n-2)! with the denominator n! = n · (n−1) · (n−2)! ↑ this part is (n−2)! and cancels with the denominator
Practice★☆☆Not started
Simplify the algebraic expression $\frac{(n+1)!}{n!}$, giving the result in terms of $n$.

Simplify the algebraic expression $\frac{(n+1)!}{n!}$, giving the result in terms of $n$.

Solve
Practice★★☆Not started
Simplify the algebraic expression $\frac{n!}{(n-2)!}$ as far as possible, expanding the numerator until terms can be cancelled with the denominator.

Simplify the algebraic expression $\frac{n!}{(n-2)!}$ as far as possible, expanding the numerator until terms can be cancelled with the denominator.

Solve

Equations with factorials

If an equation contains factorials of an unknown, the usual approach is to simplify first (as in the previous section) so that you are left with an ordinary equation.

Diagram of the process for solving an equation with factorials by algebraic simplification and checking for natural numbers x!=120 try small values 3! = 64! = 24 5! = 120 ✓
Practice★☆☆Not started
Determine the value of the natural number $x$ given that $x! = 120$.

Determine the value of the natural number $x$ given that $x! = 120$.

Solve
Practice★★☆Not started
Solve the equation $\frac{(x+2)!}{(x+1)!} = 15$, stating the value of $x$.

Solve the equation $\frac{(x+2)!}{(x+1)!} = 15$, stating the value of $x$.

Solve

Summary

See all the exercises in this section