4.1 Factorials, Binomial Coefficients and (1 + b)ⁿ
SL 1.9
- the factorial notation \(n!\)
- the binomial coefficient \(\binom{n}{r}\) (also written \({}^n\mathrm{C}_r\))
- Pascal’s triangle and its patterns
- how to expand \((1 + b)^n\), and how to use the expansion for products and approximations
The notation \(n!\)
Factorial Not in the formula booklet
For a positive integer \(n\),
\[ n! = n \times (n - 1) \times (n - 2) \times \cdots \times 3 \times 2 \times 1 . \]
We also define \(0! = 1\).
Read \(n!\) as “\(n\) factorial”. For example \(4! = 4 \times 3 \times 2 \times 1 = 24\).
A very useful fact: \(n! = n \times (n - 1)!\). For example \(6! = 6 \times 5!\). This lets you cancel factorials without multiplying everything out.
Example 1 (a) Without a calculator, find \(5! - 4!\).
(b) Simplify \(\dfrac{(n + 1)!}{(n - 1)!}\).
(a) \(5! - 4! = 5 \times 4! - 4! = 4 \times 4! = 4 \times 24 = 96\).
(b) \((n + 1)! = (n + 1) \times n \times (n - 1)!\), so
\[ \frac{(n + 1)!}{(n - 1)!} = \frac{(n + 1)\,n\,(n - 1)!}{(n - 1)!} = n(n + 1) . \]
The notation \(\binom{n}{r}\)
Binomial coefficient In the formula booklet
For integers \(0 \le r \le n\),
\[ {}^n\mathrm{C}_r = \frac{n!}{r!(n - r)!} \]
It is also written \(\dbinom{n}{r}\).
- \(n\): the number of objects to choose from (a positive integer, or \(0\))
- \(r\): the number of objects chosen (\(0 \le r \le n\))
- \(!\): factorial, for example \(r! = r \times (r - 1) \times \cdots \times 2 \times 1\)
Read \(\binom{n}{r}\) as “\(n\) choose \(r\)”. (In Unit 5 you will see that it counts the ways to choose \(r\) objects from \(n\).)
Cancelling gives a quick way to work it out by hand: \(r\) numbers counting down from \(n\), divided by \(r!\).
\[ \binom{9}{3} = \frac{9!}{3!\,6!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \]
Two facts that follow from the definition:
\[ \binom{n}{0} = \binom{n}{n} = 1, \qquad \binom{n}{r} = \binom{n}{n - r} . \]
Example 2 (a) Evaluate \(\dbinom{10}{2}\) without a calculator.
(b) Show that \(\dbinom{15}{4} = \dfrac{15 \times 14 \times 13 \times 12}{4!}\).
(a) \(\dbinom{10}{2} = \dfrac{10 \times 9}{2 \times 1} = 45\).
(b) \(\dbinom{15}{4} = \dfrac{15!}{4!\,11!}\). Since \(15! = 15 \times 14 \times 13 \times 12 \times 11!\), the \(11!\) cancels:
\[ \binom{15}{4} = \frac{15 \times 14 \times 13 \times 12 \times 11!}{4!\,11!} = \frac{15 \times 14 \times 13 \times 12}{4!} . \]
Example 3 Find the positive integer \(n\) such that \(\dbinom{n}{3} = 4\dbinom{n}{2}\).
\[ \frac{n(n - 1)(n - 2)}{6} = 4 \times \frac{n(n - 1)}{2} \]
\(n \ge 3\), so \(n(n - 1) \neq 0\) and we can divide both sides by \(n(n - 1)\):
\[ \frac{n - 2}{6} = 2 \;\Longrightarrow\; n - 2 = 12 \;\Longrightarrow\; n = 14 . \]
Pascal’s triangle
Expand \((1 + b)^n\) for small \(n\) and write down only the coefficients:
| \(n\) | Expansion of \((1 + b)^n\) | Coefficients |
|---|---|---|
| 0 | \(1\) | 1 |
| 1 | \(1 + b\) | 1 \(\;\) 1 |
| 2 | \(1 + 2b + b^2\) | 1 \(\;\) 2 \(\;\) 1 |
| 3 | \(1 + 3b + 3b^2 + b^3\) | 1 \(\;\) 3 \(\;\) 3 \(\;\) 1 |
| 4 | \(1 + 4b + 6b^2 + 4b^3 + b^4\) | 1 \(\;\) 4 \(\;\) 6 \(\;\) 4 \(\;\) 1 |
| 5 | \(1 + 5b + 10b^2 + 10b^3 + 5b^4 + b^5\) | 1 \(\;\) 5 \(\;\) 10 \(\;\) 10 \(\;\) 5 \(\;\) 1 |
The triangle of coefficients is called Pascal’s triangle. Look at its patterns:
- Row \(n\) has \(n + 1\) numbers. The first and last are always \(1\).
- Each row is symmetric: it reads the same from both ends.
- Each number inside the triangle is the sum of the two numbers just above it. For example \(10 = 4 + 6\).
- In each term, the powers of \(b\) go up \(0, 1, 2, \ldots, n\).
The numbers in row \(n\) are exactly \(\binom{n}{0}, \binom{n}{1}, \ldots, \binom{n}{n}\). For example, row 4 is \(\binom{4}{0}, \binom{4}{1}, \binom{4}{2}, \binom{4}{3}, \binom{4}{4} = 1, 4, 6, 4, 1\).
Pascal’s triangle is fine for small \(n\). For \((1 + b)^{20}\) we would need 20 rows, so instead we use \(\binom{n}{r}\) directly.
Binomial theorem for \((1 + b)^n\), \(n \in \mathbb{Z}^+\) Not in the formula booklet
\[ (1 + b)^n = \binom{n}{0} + \binom{n}{1}b + \binom{n}{2}b^2 + \cdots + \binom{n}{r}b^r + \cdots + \binom{n}{n}b^n \]
- \(b\): any number (or expression) in the bracket
- \(n\): the power, a positive integer
- \(\binom{n}{r}\): the binomial coefficient \({}^n\mathrm{C}_r\)
- \(r\): the power of \(b\) in that term (\(r = 0, 1, 2, \ldots, n\))
It is the booklet’s binomial theorem (4.2) with \(a = 1\).
Menu → Probability → Combinations, then type nCr(9,3) → 84.
Example 4 (a) Use Pascal’s triangle to write down the first five terms of \((1 + b)^6\), in ascending powers of \(b\).
(b) Hence write down the first five terms of (i) \((1 - b)^6\)
(ii) \((1 + 3x)^6\).
(a) Row 6 is \(1, 6, 15, 20, 15, 6, 1\) (from row 5: \(1+5 = 6\), \(5 + 10 = 15\), \(10 + 10 = 20\), …).
\[ (1 + b)^6 = 1 + 6b + 15b^2 + 20b^3 + 15b^4 + \cdots \]
(b)(i) Replace \(b\) by \(-b\). Odd powers change sign:
\[ (1 - b)^6 = 1 - 6b + 15b^2 - 20b^3 + 15b^4 - \cdots \]
(ii) Replace \(b\) by \(3x\). Put brackets around \(3x\):
\[ 1 + 6(3x) + 15(3x)^2 + 20(3x)^3 + 15(3x)^4 = 1 + 18x + 135x^2 + 540x^3 + 1215x^4 + \cdots \]
Products of expansions
To expand a product up to a certain power, expand each bracket that far, then multiply, keeping only the terms you need.
Example 5 (a) Find the first three terms of \((1 + 2x)^5\) and of \((1 - x)^7\), in ascending powers of \(x\).
(b) Hence find the expansion of \((1 + 2x)^5(1 - x)^7\) up to the term in \(x^2\).
(a)
\[ (1 + 2x)^5 = 1 + 5(2x) + 10(2x)^2 + \cdots = 1 + 10x + 40x^2 + \cdots \]
\[ (1 - x)^7 = 1 + 7(-x) + 21(-x)^2 + \cdots = 1 - 7x + 21x^2 + \cdots \]
(b) Multiply, but keep only the terms up to \(x^2\):
\[ \begin{aligned} (1 + 10x + 40x^2)(1 - 7x + 21x^2) &= 1 - 7x + 21x^2 + 10x - 70x^2 + 40x^2 + \cdots \\ &= 1 + 3x - 9x^2 + \cdots \end{aligned} \]
Approximations
When \(x\) is small, the higher powers \(x^3, x^4, \ldots\) are very small. So the first few terms give a good approximation.
Example 6 (a) Find the first four terms of \((1 + 2x)^6\) in ascending powers of \(x\).
(b) By choosing a suitable value of \(x\), use your answer to estimate \(1.02^6\).
(a) \(1 + 6(2x) + 15(2x)^2 + 20(2x)^3 = 1 + 12x + 60x^2 + 160x^3 + \cdots\)
(b) \(1 + 2x = 1.02\) when \(x = 0.01\):
\[ 1.02^6 \approx 1 + 0.12 + 0.006 + 0.000\,16 = 1.126\,16 . \]
(The true value is \(1.126\,162\ldots\) — the four terms are very accurate because \(x\) is small.)
Common errors
| ✗ Wrong | ✓ Right | Why |
|---|---|---|
| \((1 + 3x)^6 = 1 + 6 \cdot 3x + 15 \cdot 3x^2 + \cdots\) | \(15(3x)^2 = 135x^2\) | Keep brackets: the 3 is squared too. |
| \((1 - b)^6\) with every sign \(+\) | signs alternate | \((-b)^3 = -b^3\), \((-b)^4 = +b^4\). |
| \(\binom{9}{3} = \frac{9!}{3!}\) | \(\frac{9!}{3!\,6!}\) | Two factorials in the denominator. |
| \(0! = 0\) | \(0! = 1\) | It is defined as 1, so that \(\binom{n}{0} = 1\). |
| Multiplying all terms in a product | keep only terms up to the power asked | It saves time and avoids mistakes. |
Exercises
1 Without a calculator, evaluate
(a) \(\dfrac{8!}{6!}\)
(b) \(\dbinom{7}{3}\)
(c) \(\dbinom{12}{10}\)
(a) \(8 \times 7 = 56\)
(b) \(\dfrac{7 \times 6 \times 5}{6} = 35\)
(c) \(\dbinom{12}{2} = \dfrac{12 \times 11}{2} = 66\)
2 Simplify \(\dfrac{n!}{(n - 2)!}\).
\(\dfrac{n(n - 1)(n - 2)!}{(n - 2)!} = n(n - 1)\).
3 Solve \(\dbinom{n}{2} = 28\), where \(n\) is a positive integer.
\(\dfrac{n(n - 1)}{2} = 28 \Rightarrow n^2 - n - 56 = 0 \Rightarrow (n - 8)(n + 7) = 0\). So \(n = 8\).
4 Write down row 7 of Pascal’s triangle, and use it to expand \((1 + x)^7\).
Row 7: \(1, 7, 21, 35, 35, 21, 7, 1\).
\((1 + x)^7 = 1 + 7x + 21x^2 + 35x^3 + 35x^4 + 21x^5 + 7x^6 + x^7\).
5 Expand \((1 - 2x)^4\) fully.
\(1 + 4(-2x) + 6(-2x)^2 + 4(-2x)^3 + (-2x)^4 = 1 - 8x + 24x^2 - 32x^3 + 16x^4\).
6 Find the first three terms of \(\left(1 + \dfrac{x}{2}\right)^{10}\) in ascending powers of \(x\).
\(1 + 10\left(\dfrac{x}{2}\right) + 45\left(\dfrac{x}{2}\right)^2 = 1 + 5x + \dfrac{45}{4}x^2 + \cdots\)
7 Find the coefficient of \(x^3\) in \((1 + 3x)^5\).
\(\dbinom{5}{3}(3x)^3 = 10 \times 27x^3\), so the coefficient is \(270\).
8 Find the expansion of \((1 + x)^4(1 - 3x)^3\) up to the term in \(x^2\).
\((1 + 4x + 6x^2)(1 - 9x + 27x^2) = 1 - 9x + 27x^2 + 4x - 36x^2 + 6x^2 + \cdots = 1 - 5x - 3x^2 + \cdots\)
9 The coefficient of \(x^2\) in \((1 + kx)^6\) is \(60\). Find the possible values of \(k\).
\(\dbinom{6}{2}k^2 = 15k^2 = 60 \Rightarrow k^2 = 4 \Rightarrow k = \pm 2\).
10 Use the first four terms of \((1 - x)^8\) to estimate \(0.99^8\), correct to 5 decimal places.
\((1 - x)^8 \approx 1 - 8x + 28x^2 - 56x^3\). With \(x = 0.01\):
\[ 1 - 0.08 + 0.0028 - 0.000\,056 = 0.922\,744 \approx 0.922\,74 . \]
(The true value is \(0.922\,744\,69\ldots\))