This note covers the counting rules for arranging and choosing objects, including permutations and combinations, and then extends the binomial theorem to negative and fractional powers. The extended theorem gives an infinite series that you can use to approximate expressions such as square roots, provided \(x\) is small enough.
Key ideas
Multiplication principle. If one task can be done in \(m\) ways and a second, independent task can be done in \(n\) ways, then doing both can be done in \(m\times n\) ways. For three choices of starter, four of main course and two of dessert, there are \(3\times4\times2=24\) different meals.
Addition principle. If a job can be done in one of two ways that cannot happen together, add the numbers of ways.
Permutations count arrangements, where order matters. The number of ways to arrange \(n\) different objects in a line is \(n!\). The number of ways to pick and arrange \(r\) of them is \(^{n}P_{r}\). When some objects are identical, divide by the factorial of each repeated count, because swapping identical objects gives no new arrangement.
Combinations count selections, where order does not matter. The number of ways to choose \(r\) objects from \(n\) is \(\binom{n}{r}\). A useful test is to ask "would swapping two chosen items give a different outcome?" If yes, use permutations. If no, use combinations.
Restrictions. For "at least" or "not both" conditions, it is often easier to count the complement. That means taking the total and subtracting the unwanted cases.
The extended binomial theorem. The finite theorem needs \(n\) to be a positive integer. For any rational \(n\), such as \(-2\) or \(\tfrac12\), the expansion of \((1+x)^n\) never stops. It is an infinite series, and it is only valid when \(|x|\lt1\). If the bracket is not of the form \((1+x)\), factorise first: \((a+bx)^n=a^n\left(1+\dfrac{b}{a}x\right)^n\). The validity condition then becomes \(\left|\dfrac{b}{a}x\right|\lt1\).
When \(n\) is a positive integer, the series ends by itself because a factor \((n-n)\) appears. For other values the factors never reach zero, so the series continues.
How the formulas arise
To arrange 3 of 8 different runners in order, there are 8 choices for first place, then 7 for second, then 6 for third. Multiplying gives \(8\times7\times6\), which is \(\dfrac{8!}{5!}\). That is the idea behind \(^{n}P_{r}\). A combination ignores order, so each selection of \(r\) objects has been counted \(r!\) times among the permutations. Dividing by \(r!\) gives \(\dbinom{n}{r}\).
The terms of the extended series follow a clear pattern. The \(k\)th power of \(x\) has a coefficient with \(k\) factors on top, starting at \(n\) and decreasing by 1 each time, and \(k!\) underneath. Write the factors out in full before simplifying, as it makes sign errors with negative \(n\) much easier to spot.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(^{n}P_{r}=\dfrac{n!}{(n-r)!}\) | Ordered selections of \(r\) objects from \(n\) different objects. |
| \(^{n}C_{r}=\dbinom{n}{r}=\dfrac{n!}{r!(n-r)!}\) | Unordered selections of \(r\) objects from \(n\). |
| \(\dfrac{n!}{p!\,q!}\) | Arrangements of \(n\) objects when \(p\) are identical of one kind and \(q\) identical of another. |
| \((1+x)^n=1+nx+\dfrac{n(n-1)}{2!}x^2+\dfrac{n(n-1)(n-2)}{3!}x^3+\cdots\) | Extended binomial series for rational \(n\), valid for \(|x|\lt1\). |
| \((a+bx)^n=a^n\left(1+\dfrac{b}{a}x\right)^n\) | Rewrites a general bracket so that the series can be used. |
Worked examples
Example 1 (a) How many different arrangements are there of the letters of BALLOON? (b) A team of 5 is chosen from 6 men and 4 women. How many teams contain at least 2 women?
(a) There are 7 letters. L appears twice and O appears twice, so divide by \(2!\) for each repeated letter.
Number of arrangements \(=\dfrac{7!}{2!\,2!}=\dfrac{5040}{4}=1260\).
(b) Count the complement. Total teams: \(\dbinom{10}{5}=252\).
Teams with 0 women (5 men): \(\dbinom65=6\).
Teams with exactly 1 woman (1 woman, 4 men): \(\dbinom41\dbinom64=4\times15=60\).
Teams with at least 2 women: \(252-6-60=186\).
Answer: (a) 1260 (b) 186
Example 2 Find the first four terms of the expansion of \((1+3x)^{-2}\) in ascending powers of \(x\), and state the values of \(x\) for which it is valid.
Use \((1+y)^n\) with \(y=3x\) and \(n=-2\).
\(1+(-2)(3x)+\dfrac{(-2)(-3)}{2!}(3x)^2+\dfrac{(-2)(-3)(-4)}{3!}(3x)^3\)
\(=1-6x+\dfrac{6}{2}\times 9x^2+\dfrac{-24}{6}\times27x^3\)
\(=1-6x+27x^2-108x^3\)
The series is valid when \(|3x|\lt1\), that is \(|x|\lt\dfrac13\).
Answer: \(1-6x+27x^2-108x^3+\cdots\), valid for \(|x|\lt\tfrac13\)
Example 3 (a) Expand \(\sqrt{4+x}\) as far as the term in \(x^2\) and state when the expansion is valid. (b) Use it with \(x=0.2\) to estimate \(\sqrt{4.2}\).
(a) Factorise: \(\sqrt{4+x}=(4+x)^{1/2}=4^{1/2}\left(1+\dfrac x4\right)^{1/2}=2\left(1+\dfrac x4\right)^{1/2}\).
Expand with \(n=\tfrac12\) and \(y=\tfrac x4\):
\(\left(1+\dfrac x4\right)^{1/2}=1+\dfrac12\cdot\dfrac x4+\dfrac{\frac12\left(-\frac12\right)}{2!}\left(\dfrac x4\right)^2+\cdots=1+\dfrac x8-\dfrac{x^2}{128}+\cdots\)
Multiply by 2: \(\sqrt{4+x}=2+\dfrac x4-\dfrac{x^2}{64}+\cdots\)
Valid when \(\left|\dfrac x4\right|\lt1\), that is \(|x|\lt4\).
(b) Substitute \(x=0.2\), which is inside the valid range: \(2+\dfrac{0.2}{4}-\dfrac{0.04}{64}=2+0.05-0.000625=2.049375\).
Answer: \(2+\tfrac x4-\tfrac{x^2}{64}\), valid for \(|x|\lt4\); \(\sqrt{4.2}\approx2.0494\)
Common errors
- Using permutations when order does not matter, or combinations when it does. Decide this before you pick a formula.
- Forgetting to divide by factorials for repeated letters, which overcounts arrangements.
- Applying the extended series to \((4+x)^{1/2}\) without taking out the factor of 4. The series only works from the form \((1+\cdots)\).
- Raising the factor to the power incorrectly. For \((4+x)^{1/2}\) the outside factor is \(4^{1/2}=2\), not 4 or \(\frac12\times4\).
- Leaving out the validity condition, or writing it for \(x\) when the bracket contains \(3x\) or \(\frac x4\). The condition applies to the whole term inside the bracket.
Exam tip: When a question says "at least", write down the complement first. Counting what you do not want is often one or two terms, whereas counting what you do want may take many.
Check yourself
- Eight runners compete in a race. In how many ways can the gold, silver and bronze medals be awarded?
Answer
Order matters: \(^{8}P_3=8\times7\times6=336\). - A committee of 4 is chosen from 10 people. Two of them, Ana and Ben, refuse to serve together. How many committees are possible?
Answer
Total \(\dbinom{10}{4}=210\). Committees with both Ana and Ben: choose the other 2 from 8, so \(\dbinom82=28\). Allowed committees: \(210-28=182\). - Find the first three terms of \((1-2x)^{-1/2}\) in ascending powers of \(x\).
Answer
With \(y=-2x\) and \(n=-\tfrac12\): \(1+\left(-\tfrac12\right)(-2x)+\dfrac{\left(-\frac12\right)\left(-\frac32\right)}{2}(-2x)^2=1+x+\dfrac38\times4x^2=1+x+\tfrac32x^2\). - State the values of \(x\) for which the expansion of \((2+5x)^{-3}\) is valid.
Answer
\((2+5x)^{-3}=2^{-3}\left(1+\tfrac52x\right)^{-3}\). The series needs \(\left|\tfrac52x\right|\lt1\), so \(|x|\lt\tfrac25\).