Factorials

The factorial $n!$ is shorthand for "$n$ multiplied by $n - 1$ multiplied by $n - 2$ multiplied by $\dots$ multiplied by $1$". It is the most important single notation in combinatorics , and a tool you will use in algebra, calculus, and probability.

Definitions

For a positive integer $n$: $n! = n \cdot (n - 1) \cdot (n - 2) \cdots 3 \cdot 2 \cdot 1.$

By convention: $0! = 1.$ (The empty product is $1$, just as the empty sum is $0$. We will see in a moment why this convention is forced on us.)

The first few values:

Growth is super-exponential: $10! \approx 3.6 \times 10^6$ but $20! \approx 2.4 \times 10^{18}$.

Recursive relation

$n! = n \cdot (n - 1)! \quad \text{for } n \ge 1.$

This is the most powerful identity: $5! = 5 \cdot 4!$, $7! = 7 \cdot 6 \cdot 5!$, etc. Use it to simplify every ratio of factorials.

Setting $n = 1$: $1! = 1 \cdot 0!$, so $0! = 1$ is forced by consistency.

Simplifying ratios

The single most useful technique: when computing $\dfrac{m!}{n!}$ with $m > n$, write $\dfrac{m!}{n!} = m \cdot (m - 1) \cdots (n + 1).$

So $\dfrac{10!}{7!} = 10 \cdot 9 \cdot 8 = 720$. Never compute both factorials in full first.

Equations involving factorials

Some problems require solving for $n$ in a factorial equation. The standard moves: express factorials using the recursion, divide common factors, end up with a polynomial.

Worked examples

Example 1. Compute $\dfrac{12!}{10! \cdot 2!}$.

$\dfrac{12 \cdot 11 \cdot 10!}{10! \cdot 2!} = \dfrac{12 \cdot 11}{2} = 66$.

Example 2. Compute $\dfrac{8! - 7!}{6!}$.

$\dfrac{8 \cdot 7 \cdot 6! - 7 \cdot 6!}{6!} = 8 \cdot 7 - 7 = 56 - 7 = 49$.

Example 3. Find $n$ if $(n + 1)! = 12 \cdot (n - 1)!$.

$(n + 1)! = (n + 1) \cdot n \cdot (n - 1)!$, so $(n + 1) \cdot n = 12$, i.e. $n^2 + n - 12 = 0$, giving $n = 3$ (rejecting $n = -4$).

Example 4. Find $n$ such that $\dfrac{1}{8!} + \dfrac{1}{9!} = \dfrac{n}{10!}$.

LCM is $9!$ on the left: $\dfrac{9 + 1}{9!} = \dfrac{10}{9!}$. So $\dfrac{n}{10!} = \dfrac{10}{9!} = \dfrac{10 \cdot 10}{10!} = \dfrac{100}{10!}$. Thus $n = 100$.

Example 5 (harder). Show that $n! \cdot (n + 2) = (n + 2)! - (n + 1)!$.

RHS $= (n + 2) \cdot (n + 1)! - (n + 1)! = (n + 1)!\,(n + 2 - 1) = (n + 1)!\,(n + 1) = (n + 1) \cdot (n + 1)!$ ... wait, let me redo.

RHS $= (n + 2)! - (n + 1)! = (n + 1)!\,[(n + 2) - 1] = (n + 1)! \cdot (n + 1)$.

LHS $= n! \cdot (n + 2)$.

These are not equal in general. Let me re-state the identity I meant: $n! \cdot (n + 1) = (n + 1)!$ , trivial.

Try: $(n + 2)! - 2(n + 1)! + n! = ?$ Let's verify with $n = 3$: $5! - 2 \cdot 4! + 3! = 120 - 48 + 6 = 78$. And $n!(n^2 + 2n + 2 - 2n - 2) = n! \cdot n^2 = 6 \cdot 9 = 54$ , that does not match.

Let me give a cleaner Example 5:

Example 5 (harder, redone). Prove $\dfrac{1}{r!} - \dfrac{1}{(r + 1)!} = \dfrac{r}{(r + 1)!}$ for $r \ge 0$.

LHS $= \dfrac{(r + 1) - 1}{(r + 1)!} = \dfrac{r}{(r + 1)!}$. $\qed$

This identity is used in telescoping sums , a key technique in Class-XII series problems.

Try it yourself

1. Compute $\dfrac{10!}{8!}$, $\dfrac{15!}{13! \cdot 2!}$, $\dfrac{20!}{18! \cdot 2!}$.
2. Compute $\dfrac{6!}{4! - 3!}$. (Hint: simplify denominator first.)
3. Find $n$ if $\dfrac{n!}{2 \cdot (n - 2)!} = 15$.
4. Show $\dfrac{(n + 1)!}{n!} = n + 1$.
5. Find $n$ if $(n + 2)! = 60 \cdot (n - 1)!$.
6. Simplify $\dfrac{n!}{(n - r)!}$ as a product.
7. Compute $0!, 1!, 2!, \dots, 10!$ from scratch using the recursion.
8. Find $n$: $(n + 1)! + n! = 7 \cdot (n - 1)!$.
9. Prove $n \cdot n! = (n + 1)! - n!$.
10. Compute $\dfrac{12!}{4! \cdot 8!}$.
11. Show $\dfrac{1}{k!\,(n - k)!} = \dfrac{1}{n!} \cdot \binom{n}{k}$.
12. Compute $\dfrac{(2n)!}{(2n - 2)! \cdot 2!}$ in terms of $n$.

Pitfalls / Tricks

• $0! = 1$. Memorise; do not "calculate" it.
• Always simplify before computing. $\dfrac{100!}{99!} = 100$, not "compute $100!$ then divide".
• Factorials grow very fast. By $n = 13$, $n!$ exceeds $10^{9}$.
• Insight. Recognising $\dfrac{n!}{(n - r)!}$ as a falling product $n(n-1)\cdots(n-r+1)$ is the master simplification. Apply it whenever you see a factorial ratio.