1. Pascal's Triangle & Binomial Coefficients

Video coming soon4 worked examples

Theory

Writing out $(x+y)^n$ for $n = 0,1,2,3,\dots$ and recording only the coefficients produces Pascal's triangle:

\begin{array}{c} 1 \\ 1 \quad 1 \\ 1 \quad 2 \quad 1 \\ 1 \quad 3 \quad 3 \quad 1 \\ 1 \quad 4 \quad 6 \quad 4 \quad 1 \\ 1 \quad 5 \quad 10 \quad 10 \quad 5 \quad 1 \end{array}

Every row starts and ends with $1$ , and each entry inside a row is the sum of the two entries above it. The triangle is also symmetrical about its centre.

Rather than extend the triangle line by line, we use factorials. For $n \in \mathbb{N}$ , $n! = n(n-1)(n-2)\cdots 3 \times 2 \times 1$ , with the special definition $0! = 1$ . The binomial coefficient (read “ $n$ choose $r$ ”) is then:

\binom{n}{r} = \frac{n!}{r!\,(n-r)!}, \qquad 0 \le r \le n

These are exactly the entries of Pascal's triangle: row $n$ is $\binom{n}{0}, \binom{n}{1}, \dots, \binom{n}{n}$ . Two properties follow, and you are expected to be able to prove both:

\textbf{Symmetry:}\quad \binom{n}{r} = \binom{n}{n-r}

\textbf{Pascal's rule:}\quad \binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}

Symmetry is the mirror image of the triangle; Pascal's rule is the “add the two above” construction written algebraically.

The Golden Rule: to prove an identity in binomial coefficients, replace every $\binom{n}{r}$ by its factorial definition, put the terms over a common denominator, and simplify until the required form appears. Never argue from a few numerical rows — that is a check, not a proof.

⚠️ Common Examiner Traps

Forgetting $0! = 1$ : this is a definition, not a calculation, and it is what makes $\binom{n}{0} = 1$ work.
Verifying instead of proving: checking an identity for $n=5,\ r=2$ earns no marks. A proof must hold for general $n$ and $r$ .
The common denominator: in Pascal's rule the two denominators are $r!(n-r)!$ and $(r+1)!(n-r-1)!$ . Use $(r+1)! = (r+1)\,r!$ and $(n-r)! = (n-r)(n-r-1)!$ to bridge them.
Discarding invalid solutions: when solving for $n$ , only whole-number values with $n \ge r$ are admissible.

Worked examples

Example 1

Evaluate $\binom{9}{4}$ and $\binom{12}{8}$ , and state which other coefficient equals $\binom{12}{8}$ .

Step 1: Apply the definition to $\binom{9}{4}$ :

\binom{9}{4} = \frac{9!}{4!\,5!} = \frac{362880}{24 \times 120} = \frac{362880}{2880} = 126

Step 2: For $\binom{12}{8}$ , cancelling $8!$ first keeps the numbers manageable:

\binom{12}{8} = \frac{12!}{8!\,4!} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = \frac{11880}{24} = 495

Step 3: By the symmetry property, $\binom{12}{8} = \binom{12}{12-8} = \binom{12}{4}$ , which is also $495$ — and, as Step 2 shows, is the quicker one to compute.

Example 2

Prove that $\binom{n}{r} = \binom{n}{n-r}$ .

Step 1: Start from the right-hand side and use the definition, replacing $r$ by $n-r$ :

\binom{n}{n-r} = \frac{n!}{(n-r)!\,\bigl(n-(n-r)\bigr)!}

Step 2: Simplify the second factorial in the denominator:

n - (n-r) = r \quad\implies\quad \binom{n}{n-r} = \frac{n!}{(n-r)!\,r!}

Step 3: Multiplication is commutative, so $(n-r)!\,r! = r!\,(n-r)!$ :

\binom{n}{n-r} = \frac{n!}{r!\,(n-r)!} = \binom{n}{r}

Hence the two coefficients are equal for all $0 \le r \le n$ , as required.

Example 3

Prove Pascal's rule: $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}$ .

Step 1: Write both terms using the definition:

\binom{n}{r} + \binom{n}{r+1} = \frac{n!}{r!\,(n-r)!} + \frac{n!}{(r+1)!\,(n-r-1)!}

Step 2: The common denominator is $(r+1)!\,(n-r)!$ . Multiply the first fraction by $\dfrac{r+1}{r+1}$ and the second by $\dfrac{n-r}{n-r}$ , using $(r+1)! = (r+1)r!$ and $(n-r)! = (n-r)(n-r-1)!$ :

= \frac{n!\,(r+1)}{(r+1)!\,(n-r)!} + \frac{n!\,(n-r)}{(r+1)!\,(n-r)!}

Step 3: Combine over the single denominator and simplify the numerator:

= \frac{n!\bigl[(r+1) + (n-r)\bigr]}{(r+1)!\,(n-r)!} = \frac{n!\,(n+1)}{(r+1)!\,(n-r)!} = \frac{(n+1)!}{(r+1)!\,(n-r)!}

Step 4: Compare with the definition of the target coefficient:

\binom{n+1}{r+1} = \frac{(n+1)!}{(r+1)!\,\bigl((n+1)-(r+1)\bigr)!} = \frac{(n+1)!}{(r+1)!\,(n-r)!}

The two expressions are identical, so the identity holds.

Example 4

Find the value of $n$ for which $\binom{n}{2} = 78$ .

Step 1: Expand the coefficient using the definition, cancelling $(n-2)!$ :

\binom{n}{2} = \frac{n!}{2!\,(n-2)!} = \frac{n(n-1)}{2}

Step 2: Set this equal to $78$ and rearrange into a quadratic:

\frac{n(n-1)}{2} = 78 \implies n^2 - n - 156 = 0

Step 3: Factorise and solve:

(n-13)(n+12) = 0 \implies n = 13 \ \text{ or } \ n = -12

Step 4: Since $n$ must be a positive whole number, $n = -12$ is rejected. Therefore $n = 13$ , which checks out: $\frac{13 \times 12}{2} = 78$ .