1.10 Counting principles and extended binomial theorem (IB Maths: Analysis and Approaches HL): Revision notes

Section 1

The counting principles

The multiplication principle: if one choice can be made in $m$ ways and a second, independent choice in $n$ ways, the two together can be made in $m\times n$ ways ("and" means multiply).

The addition principle: if a selection falls into separate cases that cannot happen together, add the numbers of ways for each case ("or" means add).

A useful trick is the complement: the number with "at least one" of something $=$ total $-$ number with none. For example, teams of 4 from 7 Year 12 and 5 Year 13 students with at least one Year 13 student: $\binom{12}{4} - \binom{7}{4} = 495 - 35 = 460$ .

Key termsmultiplication principleaddition principlecomplement

Exam tip

When there is a restriction (a digit must be odd, two people must sit together), deal with the restricted position or group first.

Section 2

Permutations: when order matters

A permutation is an arrangement where order matters. The number of ways to arrange $n$ different objects in a row is $n!$ .

The number of ordered selections of $r$ objects from $n$ different objects is ${}^{n}P_r = \frac{n!}{(n-r)!} = n(n-1)\cdots(n-r+1).$ For example, 4-digit PINs using different digits from 1–9: ${}^9P_4 = 9\times8\times7\times6 = 3024$ .

Objects that must be together: treat them as a single block, arrange the blocks, then arrange inside the block. Three maths books together among 7 different books: $5!\times3! = 720$ .

Key termspermutationfactorial

Common mistake

Not required by the syllabus: arrangements with identical objects and circular arrangements. Every object in an IB question here will be different and in a line.

Section 3

Combinations: when order does not matter

A combination is a selection where order does not matter. The number of ways to choose $r$ objects from $n$ different objects is ${}^{n}C_r = \binom{n}{r} = \frac{n!}{r!(n-r)!}.$ Each combination of $r$ objects corresponds to $r!$ permutations, so $\binom{n}{r} = \frac{{}^nP_r}{r!}$ .

Choosing from two groups: exactly 2 from 7 and 2 from 5 is $\binom{7}{2}\times\binom{5}{2} = 210$ (multiply, because every pair from one group goes with every pair from the other).

A neat link: the number of 4-digit PINs from 1–9 with digits in increasing order is $\binom{9}{4} = 126$ , since each set of four digits has exactly one increasing order.

Key termscombinationbinomial coefficient

Common mistake

Using ${}^nP_r$ for a committee or team. If swapping two chosen people gives the same selection, use $\binom{n}{r}$ .

Section 4

Extending the binomial theorem to any rational power

For $n \in \mathbb{Q}$ (including negative and fractional values) and $|x| < 1$ , $(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots$ This is given in the formula booklet. When $n$ is not a positive integer the series is infinite, and it only converges when $|x| < 1$ .

Example: $(1 - 2x)^{-3} = 1 + (-3)(-2x) + \frac{(-3)(-4)}{2}(-2x)^2 + \cdots = 1 + 6x + 24x^2 + \cdots$ , valid for $|2x| < 1$ , i.e. $|x| < \frac12$ .

Key termsextended binomial theoreminterval of validity

Common mistake

Replacing $x$ by $-2x$ but forgetting to square the whole of it: $(-2x)^2 = 4x^2$ , not $-2x^2$ or $-4x^2$ .

Exam tip

Put the substituted term in brackets every time: $\left(-2x\right)$ , $\left(\frac{3x}{8}\right)^2$ .

Section 5

Expanding (a + b)^n by first taking out a factor

The expansion only works directly for $(1 + \ldots)^n$ . For $(a + b)^n$ , first write $(a + b)^n = a^n\left(1 + \frac{b}{a}\right)^n,$ which is valid for $\left|\frac{b}{a}\right| < 1$ .

Example: $(8 + 3x)^{\frac13} = 8^{\frac13}\left(1 + \frac{3x}{8}\right)^{\frac13} = 2\left(1 + \frac{x}{8} - \frac{x^2}{64} + \cdots\right) = 2 + \frac{x}{4} - \frac{x^2}{32} + \cdots$ , valid for $|x| < \frac{8}{3}$ .

Approximations: substitute a small value of $x$ into the first few terms. With $x = 0.01$ in $(1+4x)^{-\frac12} \approx 1 - 2x + 6x^2$ you get $\frac{1}{\sqrt{1.04}} \approx 0.9806$ . Multiplying a series by a polynomial such as $(1 + kx)$ does not change its interval of validity.

Key termsfactorising out a^n

Common mistake

Forgetting to raise the factor to the power: $(8 + 3x)^{\frac13} = 2(1 + \ldots)^{\frac13}$ , not $8(1 + \ldots)^{\frac13}$ . Then forgetting to multiply every term by 2.

Must know

"And" → multiply; "or" (separate cases) → add; "at least one" → total minus none.
Order matters → ${}^nP_r$ ; order does not matter → $\binom{n}{r}$ .
Not examined: identical objects, circular arrangements, proof of the binomial theorem.
$(1+x)^n$ for $n \in \mathbb{Q}$ is an infinite series valid for $|x| < 1$ .
$(a+b)^n = a^n\left(1+\frac{b}{a}\right)^n$ , valid for $|b| < |a|$ .

That's the notes covered.

Carry on to the next subtopic.