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Mathematics AA · HL · Unit 1: Number and Algebra

Unit 1Topic 1.10Key Concepts50 min read

Counting principles and the extended binomial theorem

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Fundamental counting principle

Counting problems ask how many valid outcomes are possible without listing every outcome. The key first step is to split the process into clear stages and decide whether the available cases happen together or as alternatives.

If a process has m1m_1 choices at stage 1, m2m_2 choices at stage 2, and so on, with every choice at one stage compatible with every choice at the next, then the total number of outcomes is m1m2⋯mkm_1m_2\cdots m_k. This is the fundamental counting principle, or product rule.

Use the sum rule for mutually exclusive alternatives. If one valid case produces aa outcomes and a different, non-overlapping case produces bb outcomes, the total is a+ba+b.

A tree diagram is a useful beginner check: each complete route from the start to an endpoint represents one outcome. Once the structure is understood, multiplication is faster than drawing every branch.

Three shirt choices branching to two trouser choices and six final outfits.Three choices at the first stage and two at the second produce 3×2=63\times2=6 complete outcomes.

Open graph full size (opens in a new tab)

Key terms

Outcome
One complete possible result of a counting process.
Stage
One decision or step in a multi-step process.
Product rule
Multiply the number of choices at successive compatible stages.
Sum rule
Add the counts from mutually exclusive cases.
Mutually exclusive cases
Cases that cannot occur at the same time, so no outcome is counted twice.

Worked examples

Build an outfit

A student has 3 shirts and 2 pairs of trousers. How many different outfits contain one shirt and one pair of trousers?

  1. 1Choose a shirt in 3 ways.
  2. 2For each shirt, choose trousers in 2 ways.
  3. 3Apply the product rule: 3×2=63\times2=6.

Answer: There are 6 different outfits.

Choose one transport type

There are 5 bus routes and 3 train routes from A to B. A traveler uses exactly one route. How many choices are available?

  1. 1The traveler chooses a bus route or a train route, not both.
  2. 2The two cases are mutually exclusive.
  3. 3Apply the sum rule: 5+3=85+3=8.

Answer: There are 8 route choices.

Code with independent stages

A code contains two letters followed by three digits. Repetition is allowed. How many codes exist?

  1. 1Each letter position has 26 choices.
  2. 2Each digit position has 10 choices.
  3. 3Multiply across the five stages: 262×10326^2\times10^3.

Answer: There are 676000676000 codes.

Rules and formulas

Fundamental counting principle

N=m1m2⋯mkN=m_1m_2\cdots m_k
NN
total number of complete outcomes
mim_i
number of available choices at stage i

Sum rule for disjoint cases

N=N1+N2+⋯+NkN=N_1+N_2+\cdots+N_k
NiN_i
number of outcomes in mutually exclusive case i

Factorials

For a positive integer nn, the factorial n!n! is the product of all positive integers from nn down to 1. Factorials arise whenever distinct objects are placed into distinct positions.

By definition, 0!=10!=1. This convention makes permutation, combination, and binomial-coefficient formulas work correctly at their endpoints.

Factorials grow extremely quickly. Simplify ratios before evaluating: for example, 12!10!=12×11\frac{12!}{10!}=12\times11, not two enormous calculator values.

Equations involving factorials often simplify by cancellation. Always state the integer restriction: factorial notation in this course normally requires n∈{0,1,2,…}n\in\{0,1,2,\ldots\}.

Key terms

Factorial
The product n(n−1)(n−2)⋯2⋅1n(n-1)(n-2)\cdots2\cdot1, written n!n!.
Endpoint convention
The definition 0!=10!=1, needed for consistent counting formulas.
Cancellation
Removing common factorial factors before numerical evaluation.

Worked examples

Evaluate a factorial

Find 6!6!.

  1. 1Write 6!=6×5×4×3×2×16!=6\times5\times4\times3\times2\times1.
  2. 2Multiply carefully.

Answer: 6!=7206!=720.

Simplify a factorial ratio

Simplify 15!12!\frac{15!}{12!}.

  1. 1Write 15!=15×14×13×12!15!=15\times14\times13\times12!.
  2. 2Cancel 12!12!.
  3. 3Calculate 15×14×1315\times14\times13.

Answer: 15!12!=2730\frac{15!}{12!}=2730.

Solve a factorial equation

Solve n!(n−2)!=72\frac{n!}{(n-2)!}=72 for a positive integer nn.

  1. 1Cancel to obtain n(n−1)=72n(n-1)=72.
  2. 2Solve n2−n−72=0n^2-n-72=0, giving (n−9)(n+8)=0(n-9)(n+8)=0.
  3. 3Use the positive-integer restriction.

Answer: n=9n=9.

Rules and formulas

Factorial definition

n!=n(n−1)(n−2)⋯2⋅1n!=n(n-1)(n-2)\cdots2\cdot1
nn
a positive integer

Zero factorial

0!=10!=1

Useful cancellation

n!(n−r)!=n(n−1)⋯(n−r+1)\frac{n!}{(n-r)!}=n(n-1)\cdots(n-r+1)
rr
number of descending factors

Permutations: order matters

A permutation is an ordered arrangement. Changing the order creates a different outcome, so ABC and BAC are distinct permutations.

Arranging all nn distinct objects gives n!n! outcomes. Arranging rr objects selected from nn distinct objects gives nPr=n!(n−r)!{}_{n}P_r=\frac{n!}{(n-r)!}.

The product rule explains the formula: there are n choices for the first position, n − 1 for the second, and so on until r positions are filled.

Use permutations for ranked places, assigned roles, ordered codes without repetition, or arrangements in a line.

Key terms

Permutation
An ordered selection or arrangement.
Distinct objects
Objects regarded as different from one another.
Without replacement
A selected object cannot be selected again, reducing the next stage count.

Worked examples

Award three ranked medals

Eight runners compete for gold, silver, and bronze. In how many ways can the medals be awarded?

  1. 1The positions are different, so order matters.
  2. 2Use 8P3=8×7×6{}_{8}P_3=8\times7\times6.

Answer: There are 336 possible medal outcomes.

Arrange all objects

How many linear arrangements are possible for 7 different books?

  1. 1There are 7 choices for the first position, then 6, and so on.
  2. 2The total is 7!7!.

Answer: There are 5040 arrangements.

Recognize an ordered role selection

A club of 12 students selects a president, vice-president, and treasurer. How many outcomes are possible?

  1. 1The roles are distinct, so changing the student assigned to a role changes the outcome.
  2. 2Use 12P3{}_{12}P_3.

Answer: 12P3=12×11×10=1320{}_{12}P_3=12\times11\times10=1320.

Rules and formulas

Arrange all distinct objects

N=n!N=n!
nn
number of distinct objects

Permutations of r from n

nPr=n!(n−r)!{}_{n}P_r=\frac{n!}{(n-r)!}
nn
number of available distinct objects
rr
number of ordered positions filled

Combinations: order does not matter

A combination is an unordered selection. The group {A, B, C} is the same group as {C, A, B}, so different internal orders must not be counted again.

The number of ways to select rr objects from nn distinct objects is (nr)=n!r!(n−r)!\binom{n}{r}=\frac{n!}{r!(n-r)!}. Dividing by r!r! removes the r!r! orders of each selected group.

The symmetry identity (nr)=(nn−r)\binom{n}{r}=\binom{n}{n-r} reflects the equivalence of choosing the rr included objects or the n−rn-r excluded objects.

Use combinations for committees, teams, hands of cards, or any selection where the order of the chosen members has no meaning.

Key terms

Combination
An unordered selection.
Selection
A chosen subset of the available objects.
Symmetry
The identity (nr)=(nn−r)\binom{n}{r}=\binom{n}{n-r}.

Worked examples

Choose a committee

How many committees of 4 can be chosen from 12 students?

  1. 1The order of committee members does not matter.
  2. 2Use (124)\binom{12}{4}.

Answer: (124)=495\binom{12}{4}=495.

Select then assign roles

From 10 students, choose a team of 4 and then choose its captain. How many outcomes are possible?

  1. 1Choose the team in (104)\binom{10}{4} ways.
  2. 2For each team, choose its captain in 4 ways.
  3. 3Multiply the stages.

Answer: (104)×4=840\binom{10}{4}\times4=840.

Use symmetry efficiently

Evaluate (2018)\binom{20}{18}.

  1. 1Use (2018)=(202)\binom{20}{18}=\binom{20}{2}.
  2. 2Calculate 20×192\frac{20\times19}{2}.

Answer: (2018)=190\binom{20}{18}=190.

Rules and formulas

Combination formula

(nr)=nCr=n!r!(n−r)!\binom{n}{r}={}_{n}C_r=\frac{n!}{r!(n-r)!}
nn
number of available distinct objects
rr
number selected

Combination symmetry

(nr)=(nn−r)\binom{n}{r}=\binom{n}{n-r}

Permutation-combination link

nPr=r!(nr){}_{n}P_r=r!\binom{n}{r}

Restrictions and repeated objects

Restrictions change the sample space. Translate each condition before choosing a formula: ‘together’, ‘not together’, ‘at least’, ‘exactly’, ‘no repetition’, and ‘repetition allowed’ lead to different structures.

For objects that must stay together, treat the required group as one block, arrange the block with the remaining objects, then arrange the objects inside the block.

For objects that must not be together, the complement method is often efficient: count all arrangements and subtract those in which the forbidden objects form a block.

If nn positions contain groups of identical objects of sizes n1,n2,…,nkn_1,n_2,\ldots,n_k, the number of distinguishable arrangements is n!n1!n2!⋯nk!\frac{n!}{n_1!n_2!\cdots n_k!}, where n1+⋯+nk=nn_1+\cdots+n_k=n.

Do not confuse identical objects with repetition allowed. A length-rr code chosen from kk symbols with reuse allowed has krk^r outcomes; arranging a fixed multiset requires division by repeated factorials.

Key terms

Block method
Treating objects required to be together as one temporary object.
Complement method
Counting all outcomes and subtracting the forbidden outcomes.
Repeated objects
Identical objects whose interchange does not create a new arrangement.
Repetition allowed
A symbol may be selected again at a later stage.
Case split
Dividing a problem into mutually exclusive cases and adding their counts.

Worked examples

Two people must sit together

Seven distinct people stand in a line. In how many arrangements are A and B adjacent?

  1. 1Treat A and B as one block, giving 6 objects to arrange.
  2. 2Arrange the 6 objects in 6!6! ways.
  3. 3Arrange A and B inside their block in 2 ways.

Answer: The number is 2×6!=14402\times6!=1440.

Two people must not sit together

Seven distinct people stand in a line. In how many arrangements are A and B not adjacent?

  1. 1Count all arrangements: 7!7!.
  2. 2Subtract the adjacent arrangements: 2×6!2\times6!.
  3. 3Calculate 7!−2×6!7!-2\times6!.

Answer: There are 3600 valid arrangements.

Arrange repeated letters

How many distinguishable arrangements are there of the letters in STATISTICS?

  1. 1There are 10 letters: S appears 3 times, T appears 3 times, and I appears 2 times.
  2. 2Start with 10!10! and divide by 3!3!2!3!3!2!.

Answer: 10!3!3!2!=50400\frac{10!}{3!3!2!}=50400.

At least three women

A committee of 5 is chosen from 8 women and 6 men. How many committees contain at least 3 women?

  1. 1Split into exactly 3, exactly 4, and exactly 5 women.
  2. 2Count (83)(62)+(84)(61)+(85)(60)\binom{8}{3}\binom{6}{2}+\binom{8}{4}\binom{6}{1}+\binom{8}{5}\binom{6}{0}.
  3. 3Add the mutually exclusive cases.

Answer: There are 1316 committees.

Rules and formulas

Arrangements with identical objects

N=n!n1!n2!⋯nk!N=\frac{n!}{n_1!n_2!\cdots n_k!}
nin_i
multiplicity of identical object type i

Sequences with repetition allowed

N=krN=k^r
kk
number of available symbols
rr
number of positions

Complement count

N(valid)=N(all)−N(forbidden)N(\text{valid})=N(\text{all})-N(\text{forbidden})

Binomial coefficients and finite theorem review

The coefficient (nr)\binom{n}{r} counts the ways to choose the rr positions that receive the second component bb when expanding (a+b)n(a+b)^n. This connects combinations directly to algebra.

For n∈{0,1,2,…}n\in\{0,1,2,\ldots\}, the finite binomial theorem is (a+b)n=∑r=0n(nr)an−rbr(a+b)^n=\sum\limits_{r=0}^{n}\binom{n}{r}a^{n-r}b^r. It contains exactly n+1n+1 terms.

The general term is Tr+1=(nr)an−rbrT_{r+1}=\binom{n}{r}a^{n-r}b^r. The index starts at r=0r=0, so the term containing rr is term number r+1r+1.

If the second component is negative, include its sign inside the power. This automatically produces the correct alternating sign pattern.

Key terms

Binomial coefficient
The combination number (nr)\binom{n}{r} multiplying the term indexed by r.
Finite expansion
An expansion that terminates because the exponent is a non-negative integer.
General term
A formula that produces any chosen term without expanding everything.

Worked examples

Expand a finite binomial

Expand (2x−3)4(2x-3)^4.

  1. 1Use coefficients 1, 4, 6, 4, 1.
  2. 2Use terms (4r)(2x)4−r(−3)r\binom{4}{r}(2x)^{4-r}(-3)^r.
  3. 3Simplify each coefficient and retain the alternating signs.

Answer: (2x−3)4=16x4−96x3+216x2−216x+81(2x-3)^4=16x^4-96x^3+216x^2-216x+81.

Find a finite coefficient

Find the coefficient of x3x^3 in (1+2x)7(1+2x)^7.

  1. 1The power x3x^3 occurs when r=3r=3.
  2. 2The coefficient is (73)23\binom{7}{3}2^3.

Answer: The coefficient is 280.

Use coefficient symmetry

Explain why the coefficients of x2x^2 and x6x^6 in (1+x)8(1+x)^8 are equal.

  1. 1The coefficients are (82)\binom{8}{2} and (86)\binom{8}{6}.
  2. 2Use (82)=(88−2)=(86)\binom{8}{2}=\binom{8}{8-2}=\binom{8}{6}.

Answer: Both coefficients are 28.

Rules and formulas

Finite binomial theorem

(a+b)n=∑r=0n(nr)an−rbr(a+b)^n=\sum\limits_{r=0}^{n}\binom{n}{r}a^{n-r}b^r
nn
a non-negative integer

Finite general term

Tr+1=(nr)an−rbrT_{r+1}=\binom{n}{r}a^{n-r}b^r
rr
index beginning at 0

Pascal identity

(nr)=(n−1r−1)+(n−1r)\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}

The extended binomial theorem

The extended, or generalized, binomial theorem allows an exponent α\alpha that is negative or non-integer. Instead of stopping, the expansion usually continues indefinitely.

For ∣x∣<1|x|<1, (1+x)α=∑r=0∞(αr)xr(1+x)^\alpha=\sum\limits_{r=0}^{\infty}\binom{\alpha}{r}x^r, where (αr)=α(α−1)⋯(α−r+1)r!\binom{\alpha}{r}=\frac{\alpha(\alpha-1)\cdots(\alpha-r+1)}{r!} and (α0)=1\binom{\alpha}{0}=1.

The first terms are 1+αx+α(α−1)2!x2+α(α−1)(α−2)3!x3+⋯1+\alpha x+\frac{\alpha(\alpha-1)}{2!}x^2+\frac{\alpha(\alpha-1)(\alpha-2)}{3!}x^3+\cdots. This form is often the fastest way to generate the required number of terms.

If α=n\alpha=n is a non-negative integer, the factor n−n=0n-n=0 makes all later coefficients vanish, so the generalized formula reduces to the familiar finite theorem.

Key terms

Generalized coefficient
(αr)\binom{\alpha}{r}, defined by a descending product even when α\alpha is not a non-negative integer.
Infinite series
A sum with indefinitely many terms.
Truncation
Keeping only the first required terms of an infinite expansion.
Convergence
Approach of the partial sums to a finite value.

Worked examples

Generate generalized coefficients

Find (120)\binom{\frac{1}{2}}{0}, (121)\binom{\frac{1}{2}}{1}, (122)\binom{\frac{1}{2}}{2}, and (123)\binom{\frac{1}{2}}{3}.

  1. 1(120)=1\binom{\frac{1}{2}}{0}=1.
  2. 2(121)=12\binom{\frac{1}{2}}{1}=\frac12.
  3. 3(122)=(12)(−12)2=−18\binom{\frac{1}{2}}{2}=\frac{(\frac{1}{2})(\frac{-1}{2})}{2}=-\frac18.
  4. 4(123)=(12)(−12)(−32)6=116\binom{\frac{1}{2}}{3}=\frac{(\frac{1}{2})(\frac{-1}{2})(\frac{-3}{2})}{6}=\frac1{16}.

Answer: The coefficients are 1,12,−18,1161,\frac12,-\frac18,\frac1{16}.

Expand a square root

Write the first four terms of (1+x)12(1+x)^{\frac{1}{2}}.

  1. 1Use the generalized coefficients for alpha = 12\frac{1}{2}.
  2. 2Multiply each coefficient by the matching power of x.

Answer: (1+x)12=1+12x−18x2+116x3+⋯(1+x)^{\frac{1}{2}}=1+\frac12x-\frac18x^2+\frac1{16}x^3+\cdots, valid for ∣x∣<1|x|<1.

See why an integer power terminates

Use the generalized coefficient to explain why (1+x)3(1+x)^3 has no x4x^4 term.

  1. 1The coefficient would be (34)=3⋅2⋅1⋅04!\binom{3}{4}=\frac{3\cdot2\cdot1\cdot0}{4!}.
  2. 2The descending product contains a zero.

Answer: (34)=0\binom{3}{4}=0, so the expansion terminates after x3x^3.

Rules and formulas

Generalized binomial coefficient

(αr)=α(α−1)⋯(α−r+1)r!\binom{\alpha}{r}=\frac{\alpha(\alpha-1)\cdots(\alpha-r+1)}{r!}
α\alpha
real or rational exponent
rr
non-negative integer term index

Extended binomial theorem

(1+x)α=∑r=0∞(αr)xr,∣x∣<1(1+x)^\alpha=\sum\limits_{r=0}^{\infty}\binom{\alpha}{r}x^r,\qquad |x|<1
xx
expansion variable inside the convergence interval

First four terms

(1+x)α=1+αx+α(α−1)2!x2+α(α−1)(α−2)3!x3+⋯(1+x)^\alpha=1+\alpha x+\frac{\alpha(\alpha-1)}{2!}x^2+\frac{\alpha(\alpha-1)(\alpha-2)}{3!}x^3+\cdots

Negative and fractional powers

Negative powers often generate recognizable infinite patterns. For example, (1+x)−1=1−x+x2−x3+⋯(1+x)^{-1}=1-x+x^2-x^3+\cdots, which is also an infinite geometric series.

For a positive integer mm, (−mr)=(−1)r(m+r−1r)\binom{-m}{r}=(-1)^r\binom{m+r-1}{r}. This identity explains the alternating generalized coefficients before any substitution inside (1+x)−m(1+x)^{-m}.

Fractional powers model roots. The expansion of (1+x)12(1+x)^{\frac{1}{2}} can approximate 1+x\sqrt{1+x} when ∣x∣<1|x|<1, and (1+x)−12(1+x)^{\frac{-1}{2}} can approximate 11+x\frac{1}{\sqrt{1+x}}.

For (a+bx)α(a+bx)^\alpha, factor first: (a+bx)α=aα(1+bax)α(a+bx)^\alpha=a^\alpha\left(1+\frac{b}{a}x\right)^\alpha. The convergence condition becomes ∣bax∣<1\left|\frac{b}{a}x\right|<1.

Key terms

Negative exponent
An exponent such as −m-m, representing a reciprocal power.
Fractional exponent
An exponent representing a root together with a power.
Normalization
Factoring a binomial into aα(1+u)αa^\alpha(1+u)^\alpha before using the extended theorem.

Worked examples

Negative integer power

Write the first four terms of (1+x)−2(1+x)^{-2}.

  1. 1Use α=−2\alpha=-2.
  2. 2The coefficients are 1,−2,3,−41,-2,3,-4 for powers x0,x1,x2,x3x^0,x^1,x^2,x^3.

Answer: (1+x)−2=1−2x+3x2−4x3+⋯(1+x)^{-2}=1-2x+3x^2-4x^3+\cdots, valid for ∣x∣<1|x|<1.

Substitute a linear term

Write the first four terms of (1−2x)−12(1-2x)^{\frac{-1}{2}}.

  1. 1Use α=−12\alpha=-\frac12 and substitute u=−2xu=-2x.
  2. 2Compute 1+αu+α(α−1)2u2+α(α−1)(α−2)6u31+\alpha u+\frac{\alpha(\alpha-1)}2u^2+\frac{\alpha(\alpha-1)(\alpha-2)}6u^3.
  3. 3Simplify each power of −2x-2x.

Answer: (1−2x)−12=1+x+32x2+52x3+⋯(1-2x)^{\frac{-1}{2}}=1+x+\frac32x^2+\frac52x^3+\cdots, valid for ∣x∣<12|x|<\frac12.

Factor before expanding

Find the first three terms of (4+x)−12(4+x)^{\frac{-1}{2}}.

  1. 1Write (4+x)−12=12(1+x4)−12(4+x)^{\frac{-1}{2}}=\frac12\left(1+\frac{x}{4}\right)^{\frac{-1}{2}}.
  2. 2Use (1+u)−12=1−12u+38u2+⋯(1+u)^{\frac{-1}{2}}=1-\frac12u+\frac38u^2+\cdots.
  3. 3Substitute u=x4u=\frac{x}{4} and multiply by 12\frac{1}{2}.

Answer: (4+x)−12=12−x16+3x2256+⋯(4+x)^{\frac{-1}{2}}=\frac12-\frac{x}{16}+\frac{3x^2}{256}+\cdots, valid for ∣x∣<4|x|<4.

Rules and formulas

Negative upper coefficient

(−mr)=(−1)r(m+r−1r)\binom{-m}{r}=(-1)^r\binom{m+r-1}{r}
mm
positive integer

Normalized binomial

(a+bx)α=aα(1+bax)α(a+bx)^\alpha=a^\alpha\left(1+\frac{b}{a}x\right)^\alpha
aa
non-zero constant factored from the binomial

Resulting validity condition

∣bax∣<1\left|\frac{b}{a}x\right|<1

General term and specific coefficients

For the extended expansion (1+u)α(1+u)^\alpha, the general term is Tr+1=(αr)urT_{r+1}=\binom{\alpha}{r}u^r. It works for any non-negative integer rr, even though α\alpha need not be an integer.

To find a coefficient, express u in terms of x, identify the value of r that produces the required power, and include every numerical factor from u raised to that power.

For a transformed expression (a+bx)α(a+bx)^\alpha, either normalize first or use Tr+1=aα(αr)(bax)rT_{r+1}=a^\alpha\binom{\alpha}{r}\left(\frac{b}{a}x\right)^r.

When multiplying two series, a requested coefficient may receive contributions from several pairs of terms. List all exponent pairs that add to the required power before summing.

Key terms

Coefficient extraction
Finding the numerical multiplier of a specified power without writing unnecessary terms.
Index matching
Solving for the term index r that creates the required power of x.
Contribution
One product of terms that forms the requested power when series are multiplied.

Worked examples

Coefficient in a negative-power expansion

Find the coefficient of x3x^3 in (1−2x)−2(1-2x)^{-2}.

  1. 1Use r=3r=3, α=−2\alpha=-2, and u=−2xu=-2x.
  2. 2The coefficient is (−23)(−2)3\binom{-2}{3}(-2)^3.
  3. 3Calculate (−23)=−4\binom{-2}{3}=-4.

Answer: The coefficient of x cubed is 32.

Coefficient after normalization

Find the coefficient of x2x^2 in (4+x)−12(4+x)^{\frac{-1}{2}}.

  1. 1Normalize as 12(1+x4)−12\frac12(1+\frac{x}{4})^{\frac{-1}{2}}.
  2. 2Use the r=2r=2 coefficient (−122)=38\binom{\frac{-1}{2}}{2}=\frac{3}{8}.
  3. 3Compute 12⋅38⋅142\frac12\cdot\frac38\cdot\frac1{4^2}.

Answer: The coefficient is 3256\frac{3}{256}.

Coefficient from a product

Find the coefficient of x2x^2 in (1+x)(1+x)−12(1+x)(1+x)^{\frac{-1}{2}}.

  1. 1Use (1+x)−12=1−12x+38x2+⋯(1+x)^{\frac{-1}{2}}=1-\frac12x+\frac38x^2+\cdots.
  2. 2The x2x^2 contributions are 1⋅38x21\cdot\frac38x^2 and x⋅(−12x)x\cdot(-\frac12x).
  3. 3Add 38−12\frac38-\frac12.

Answer: The coefficient is −18-\frac18.

Rules and formulas

Extended general term

Tr+1=(αr)urT_{r+1}=\binom{\alpha}{r}u^r
rr
non-negative integer index

General term after normalization

Tr+1=aα(αr)(bax)rT_{r+1}=a^\alpha\binom{\alpha}{r}\left(\frac{b}{a}x\right)^r

Validity, convergence and approximation

When α\alpha is not a non-negative integer, the extended binomial series has the standard convergence condition ∣x∣<1|x|<1. At x=1x=1 or x=−1x=-1, endpoint behavior requires separate analysis and must not be assumed from the open interval.

After substituting u=g(x)u=g(x), apply the condition to the complete inner expression: ∣g(x)∣<1|g(x)|<1. For (1−3x)12(1-3x)^{\frac{1}{2}}, this gives ∣−3x∣<1|-3x|<1, hence ∣x∣<13|x|<\frac13.

A non-negative integer exponent produces a polynomial, so its finite expansion is valid for every real x; the convergence restriction is only relevant to the genuinely infinite expansion.

Truncating a convergent series gives an approximation. The closer the inner value is to zero, the more rapidly successive powers usually shrink. Report an approximate sign and appropriate accuracy.

A formula may define a real function on a wider domain than its expansion is valid. For example, the function may exist at a point where this particular power series does not converge.

Key terms

Radius of convergence
The distance from the expansion center within which the series converges; here the standard radius is 1 in the inner variable.
Endpoint
A boundary value excluded by the strict inequality and requiring a separate test.
Partial sum
The sum of the first finitely many terms of an infinite series.
Approximation
A numerical estimate obtained by truncating a convergent expansion.

Worked examples

Find an interval of validity

State the interval of validity for the extended expansion of (1−4x)−32(1-4x)^{\frac{-3}{2}}.

  1. 1The inner variable is u=−4xu=-4x.
  2. 2Require ∣u∣<1|u|<1: ∣−4x∣<1|-4x|<1.
  3. 3Divide by 4 to obtain ∣x∣<14|x|<\frac{1}{4}.

Answer: The interval is −14<x<14-\frac14<x<\frac14.

Approximate a square root

Use three non-constant terms of (1+x)12(1+x)^{\frac{1}{2}} to approximate 1.04\sqrt{1.04}.

  1. 1Set x=0.04x=0.04.
  2. 2Use 1+12x−18x2+116x31+\frac12x-\frac18x^2+\frac1{16}x^3.
  3. 3Calculate 1+0.02−0.0002+0.0000041+0.02-0.0002+0.000004.

Answer: 1.04≈1.019804\sqrt{1.04}\approx1.019804.

Distinguish function domain from series validity

Does the series 1−x+x2−x3+⋯1-x+x^2-x^3+\cdots represent (1+x)−1(1+x)^{-1} at x=2x=2?

  1. 1The function value (1+2)−1=13(1+2)^{-1}=\frac{1}{3} exists.
  2. 2The series requires ∣x∣<1|x|<1, but ∣2∣>1|2|>1.

Answer: No. The function exists at x=2x = 2, but this series does not converge there.

Rules and formulas

Standard convergence condition

∣x∣<1|x|<1

Substituted convergence condition

∣g(x)∣<1|g(x)|<1
g(x)g(x)
complete expression replacing the standard inner variable

Truncated approximation

(1+x)α≈∑r=0k(αr)xr(1+x)^\alpha\approx\sum\limits_{r=0}^{k}\binom{\alpha}{r}x^r
kk
highest retained power

Calculator and GDC methods

Scientific calculators and GDCs usually provide factorial, nPr, and nCr commands in a probability or mathematics menu. Enter n first, select the operation, then enter r.

Use technology to evaluate large exact counts, but write the model first. An unexplained calculator output does not show whether order matters or whether restrictions were applied correctly.

For generalized coefficients with negative or fractional α\alpha, many nCr commands are unavailable. Evaluate the descending product α(α−1)⋯(α−r+1)r!\frac{\alpha(\alpha-1)\cdots(\alpha-r+1)}{r!}, or define it as a sequence on the GDC.

A CAS may display a finite expansion or a truncated series, but you must still state the order term and validity condition. Verify signs and coefficients against the first-term pattern.

For approximations, keep full calculator precision through the retained sum and round only the final answer. Compare the size of the last retained term with the required accuracy.

Key terms

nPr command
Calculator command for ordered selections without replacement.
nCr command
Calculator command for unordered selections.
CAS expansion
Computer algebra output for an algebraic expansion or series.
Order term
A statement indicating the first omitted power or the continuation of a series.

Worked examples

Check a large combination

Evaluate (308)\binom{30}{8} on a GDC.

  1. 1Enter 30.
  2. 2Choose nCr and enter 8.
  3. 3Press enter, then retain the exact integer.

Answer: (308)=5852925\binom{30}{8}=5852925.

Generate a generalized coefficient

Use a calculator to evaluate (−324)\binom{\frac{-3}{2}}{4}.

  1. 1Enter (−32)(−52)(−72)(−92)4!\frac{(\frac{-3}{2})(\frac{-5}{2})(\frac{-7}{2})(\frac{-9}{2})}{4!}.
  2. 2Use exact fraction mode if available.
  3. 3Check that four descending factors appear.

Answer: (−324)=315128\binom{\frac{-3}{2}}{4}=\frac{315}{128}.

Numerical check without replacing proof

How should a GDC be used after deriving a restricted counting result?

  1. 1Enter the final expression exactly.
  2. 2Check it is a non-negative integer.
  3. 3Estimate whether its size is plausible from the unrestricted total.

Answer: Use the GDC to verify arithmetic, while keeping the case structure and restriction reasoning in the written solution.

Rules and formulas

Generalized coefficient recurrence

(αr+1)=(αr)α−rr+1\binom{\alpha}{r+1}=\binom{\alpha}{r}\frac{\alpha-r}{r+1}
rr
current coefficient index

Plausibility bound for a restriction

0≤N(valid)≤N(all)0\le N(\text{valid})\le N(\text{all})

IB HL problem-solving and exam strategy

Begin by classifying the problem: staged choices, ordered selection, unordered selection, repeated objects, restricted cases, finite binomial expansion, or infinite generalized expansion.

For counting, define what one outcome means. Then state whether order matters, whether replacement is allowed, and whether cases overlap. A correct formula applied to the wrong model earns little credit.

For an extended expansion, normalize the binomial, identify alpha and the complete inner variable u, write the general term, generate only the required powers, and state |u| < 1.

Use exact values during algebra. Fractions reveal coefficient structure and avoid rounding error. Convert to decimals only when an approximation or contextual answer is requested.

Finish with a reasonableness check: a count must be a non-negative integer no larger than the unrestricted total, and a truncated series should have decreasing term sizes near its center.

Key terms

Model
A mathematical representation of the choices, restrictions, or expansion structure.
Unrestricted total
The count before a restriction removes outcomes.
Method mark
Credit awarded for a valid mathematical process even if later arithmetic is incorrect.
Reasonableness check
A structural or numerical check that the result fits the problem.

Worked examples

Choose the correct counting tool

Match each situation to a method: a committee of 4; four ranked prizes; arrangements of LEVEL; six-digit codes with reuse.

  1. 1Committee: unordered selection.
  2. 2Prizes: ordered selection.
  3. 3LEVEL: repeated objects.
  4. 4Codes with reuse: product rule.

Answer: Use nCr; nPr; 5! divided by 2!2!; and 10 to the power 6, respectively.

Plan an extended expansion

Describe the method for expanding (9−2x)−12(9-2x)^{\frac{-1}{2}} through x2x^2.

  1. 1Factor: (9−2x)−12=13(1−2x9)−12(9-2x)^{\frac{-1}{2}}=\frac13(1-\frac{2x}{9})^{\frac{-1}{2}}.
  2. 2Use α=−12\alpha=\frac{-1}{2} and u=−2x9u=\frac{-2x}{9}.
  3. 3Retain r=0r = 0, 1, 2, multiply by one third, then state |2x9\frac{2x}{9}| < 1.

Answer: The expansion is 13+x27+x2162+⋯\frac13+\frac{x}{27}+\frac{x^2}{162}+\cdots, valid for ∣x∣<92|x|<\frac92.

Check a restricted count

A calculation gives 560 valid arrangements out of 420 unrestricted arrangements. What does this show?

  1. 1A restriction can only keep or remove outcomes.
  2. 2Therefore the valid count cannot exceed the unrestricted count.

Answer: The model or arithmetic is wrong; revisit overlapping cases, multiplication, or division by repeats.

Rules and formulas

Counting decision

order matters⇒nPr,order does not matter⇒(nr)\text{order matters}\Rightarrow {}_{n}P_r,\qquad \text{order does not matter}\Rightarrow\binom{n}{r}

Extended-expansion checklist

(a+bx)α=aα(1+u)α,u=bax,∣u∣<1(a+bx)^\alpha=a^\alpha(1+u)^\alpha,\quad u=\frac{b}{a}x,\quad |u|<1

IB-style practice questions with worked solutions

Attempt each question before revealing the steps. The set moves from direct counting and notation to restrictions, generalized coefficients, convergence, and approximation.

In an examination, write the structural expression before using technology and state exact answers unless an approximation is requested.

Key terms

Exact answer
An integer, fraction, radical, or algebraic form with no rounding.
Validity condition
The inequality defining where an infinite expansion represents the function.
Worked solution
A sequence of justified steps leading to the answer.

Worked examples

1. Product rule with repetition

A password contains three letters followed by four digits. Repetition is allowed. How many passwords are possible?

  1. 1There are 26 choices for each letter and 10 for each digit.
  2. 2Use 263×10426^3\times10^4.

Answer: There are 175760000 passwords.

2. Sum rule

A journey uses exactly one of 4 bus routes or 3 train routes. How many route choices are possible?

  1. 1Bus and train are mutually exclusive alternatives.
  2. 2Add 4+34+3.

Answer: There are 7 choices.

3. Restricted permutation using a block

Eight distinct books are arranged on a shelf. Three particular books must remain together. How many arrangements are possible?

  1. 1Treat the three books as one block, leaving 6 objects.
  2. 2Arrange the objects in 6!6! ways.
  3. 3Arrange the three books inside the block in 3!3! ways.
  4. 4Multiply.

Answer: 6!×3!=43206!\times3!=4320.

4. Combination

How many groups of 6 can be selected from 14 students?

  1. 1Order does not matter.
  2. 2Use (146)\binom{14}{6}.

Answer: (146)=3003\binom{14}{6}=3003.

5. Repeated letters

How many distinguishable arrangements are there of MISSISSIPPI?

  1. 1There are 11 letters.
  2. 2I appears 4 times, S appears 4 times, and P appears 2 times.
  3. 3Use 11!4!4!2!\frac{11!}{4!4!2!}.

Answer: There are 34650 arrangements.

6. Number with no repeated digits

How many six-digit numbers have no repeated digits?

  1. 1The first digit has 9 choices because it cannot be 0.
  2. 2The second digit has 9 remaining choices, including 0.
  3. 3The remaining positions have 8, 7, 6, and 5 choices.

Answer: 9×9×8×7×6×5=1360809\times9\times8\times7\times6\times5=136080.

7. Finite-binomial coefficient

Find the coefficient of x4x^4 in (2−x)6(2-x)^6.

  1. 1Use the term with r=4r=4: (64)26−4(−x)4\binom{6}{4}2^{6-4}(-x)^4.
  2. 2Calculate 15×4×115\times4\times1.

Answer: The coefficient is 60.

8. Fractional-power expansion

Find the first four terms of (1+3x)12(1+3x)^{\frac{1}{2}}.

  1. 1Use (1+u)12=1+12u−18u2+116u3+⋯(1+u)^{\frac{1}{2}}=1+\frac12u-\frac18u^2+\frac1{16}u^3+\cdots.
  2. 2Substitute u=3xu=3x.

Answer: (1+3x)12=1+32x−98x2+2716x3+⋯(1+3x)^{\frac{1}{2}}=1+\frac32x-\frac98x^2+\frac{27}{16}x^3+\cdots, valid for ∣x∣<13|x|<\frac13.

9. Negative-power expansion and validity

Expand (1−2x)−2(1-2x)^{-2} through x3x^3 and state its validity.

  1. 1Use (1−u)−2=1+2u+3u2+4u3+⋯(1-u)^{-2}=1+2u+3u^2+4u^3+\cdots.
  2. 2Substitute u=2xu=2x.
  3. 3Require ∣2x∣<1|2x|<1.

Answer: (1−2x)−2=1+4x+12x2+32x3+⋯(1-2x)^{-2}=1+4x+12x^2+32x^3+\cdots, valid for ∣x∣<12|x|<\frac12.

10. Generalized coefficient

Find the coefficient of x3x^3 in (1+x2)−3(1+\frac{x}{2})^{-3}.

  1. 1Use r=3r=3: (−33)(12)3\binom{-3}{3}(\frac{1}{2})^3.
  2. 2Calculate (−33)=−10\binom{-3}{3}=-10.
  3. 3Multiply by 18\frac{1}{8}.

Answer: The coefficient is −54-\frac54.

11. Approximation

Use terms through x3x^3 in (1+x)−12(1+x)^{\frac{-1}{2}} to approximate 10.96\frac{1}{\sqrt{0.96}}.

  1. 1Set x=−0.04x=-0.04.
  2. 2Use 1−12x+38x2−516x31-\frac12x+\frac38x^2-\frac5{16}x^3.
  3. 3Substitute without premature rounding.

Answer: 10.96≈1.02062\frac{1}{\sqrt{0.96}}\approx1.02062.

12. Restricted committee

A committee of 4 is chosen from 7 women and 5 men. How many committees contain exactly 2 women?

  1. 1Choose 2 women in (72)\binom{7}{2} ways.
  2. 2Choose 2 men in (52)\binom{5}{2} ways.
  3. 3Multiply the independent selections.

Answer: (72)(52)=210\binom{7}{2}\binom{5}{2}=210.

Rules and formulas

Ordered selection

nPr=n!(n−r)!{}_{n}P_r=\frac{n!}{(n-r)!}

Unordered selection

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

Extended general term

Tr+1=(αr)urT_{r+1}=\binom{\alpha}{r}u^r

Topic summary and cheat sheet

Counting starts with structure: multiply successive compatible stages, add mutually exclusive cases, and decide whether order matters.

Use nPr{}_{n}P_r for ordered selections and (nr)\binom{n}{r} for unordered selections. Divide by repeated factorials only when objects are identical.

For a non-negative integer exponent, use the finite theorem. For negative or non-integer α\alpha, normalize to (1+u)α(1+u)^\alpha, use generalized coefficients, and state ∣u∣<1|u|<1.

A strong IB solution labels restrictions and cases, preserves exact arithmetic, shows the general term when extracting a coefficient, and checks that the result is reasonable.

Key terms

Counting checklist
AND or OR? Order or no order? Replacement or no replacement? Restrictions? Identical objects?
Expansion checklist
Normalize; identify alpha and u; choose r; simplify exactly; state |u| < 1.
Final check
A count is a non-negative integer; an infinite expansion includes its validity condition and continuation.

Worked examples

One-minute method choice

Which four questions should you ask before beginning a counting calculation?

  1. 1Does order matter?
  2. 2Is repetition or replacement allowed?
  3. 3Are there restrictions or identical objects?
  4. 4Are the cases successive or mutually exclusive alternatives?

Answer: Those decisions determine whether to use a product, sum, permutation, combination, repeated-object formula, block, complement, or case split.

One-minute extended-binomial method

What is the workflow for (a+bx)α(a+bx)^\alpha when α\alpha is negative or non-integer? Assume a>0a>0 and real b,αb,\alpha.

  1. 1Factor to aα(1+u)αa^\alpha(1+u)^\alpha.
  2. 2Set u=bxau=\frac{bx}{a}.
  3. 3Use (αr)ur\binom{\alpha}{r}u^r.
  4. 4State ∣u∣<1|u|<1.

Answer: Retain only the required powers, include the outside factor, and keep exact coefficients.

Rules and formulas

Factorial

n!=n(n−1)⋯2⋅1,0!=1n!=n(n-1)\cdots2\cdot1,\qquad 0!=1

Permutation and combination

nPr=n!(n−r)!,(nr)=n!r!(n−r)!{}_{n}P_r=\frac{n!}{(n-r)!},\qquad \binom{n}{r}=\frac{n!}{r!(n-r)!}

Repeated objects

N=n!n1!n2!⋯nk!N=\frac{n!}{n_1!n_2!\cdots n_k!}

Extended binomial theorem

(1+u)α=∑r=0∞(αr)ur,∣u∣<1(1+u)^\alpha=\sum\limits_{r=0}^{\infty}\binom{\alpha}{r}u^r,\qquad |u|<1

Generalized coefficient

(αr)=α(α−1)⋯(α−r+1)r!\binom{\alpha}{r}=\frac{\alpha(\alpha-1)\cdots(\alpha-r+1)}{r!}

Topic summary

  • Multiply the numbers of choices at successive compatible stages.
  • Add counts only when cases are mutually exclusive and collectively cover the allowed outcomes.
  • Use n!=n(n−1)⋯1n!=n(n-1)\cdots1 and remember 0!=10!=1.
  • Use nPr=n!(n−r)!{}_{n}P_r=\frac{n!}{(n-r)!} when order matters.
  • Use (nr)=n!r!(n−r)!\binom{n}{r}=\frac{n!}{r!(n-r)!} when order does not matter.
  • Separate group selection from later role assignment when both occur.
  • Use blocks for objects required to stay together and complements for forbidden arrangements when efficient.
  • For identical objects, divide n!n! by the factorial of each multiplicity.
  • If repetition is allowed in each of rr positions from kk symbols, there are krk^r sequences.
  • List all exact cases for ‘at least’ and ‘at most’ restrictions before adding.
  • The finite theorem is (a+b)n=∑r=0n(nr)an−rbr(a+b)^n=\sum\limits_{r=0}^{n}\binom{n}{r}a^{n-r}b^r.
  • The finite general term is Tr+1=(nr)an−rbrT_{r+1}=\binom{n}{r}a^{n-r}b^r.
  • For arbitrary α\alpha, define (αr)=α(α−1)⋯(α−r+1)r!\binom{\alpha}{r}=\frac{\alpha(\alpha-1)\cdots(\alpha-r+1)}{r!}.
  • The extended theorem is (1+u)α=∑r=0∞(αr)ur(1+u)^\alpha=\sum\limits_{r=0}^{\infty}\binom{\alpha}{r}u^r.
  • For a genuinely infinite extended expansion, state ∣u∣<1|u|<1.
  • Normalize (a+bx)α(a+bx)^\alpha as aα(1+bxa)αa^\alpha(1+\frac{bx}{a})^\alpha.
  • When extracting a coefficient, include the full numerical factor raised to the chosen power.
  • When multiplying series, add every pair of terms that contributes to the requested power.
  • Use GDC nPr and nCr for evaluation, but show the mathematical model first.
  • For negative or fractional upper coefficients, use the descending-product definition if calculator nCr is unavailable.
  • Keep exact fractions through the expansion and round only a requested final approximation.
  • A count must be a non-negative integer and cannot exceed its unrestricted total.
  • An infinite expansion must show continuation and its validity condition.
  • Near the expansion center, decreasing omitted terms provide a useful approximation check.