Combinatorics, Series and Vectors

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Most tested topics

11. Permutations and combinations

A permutation counts arrangements, so changing the order or assigned roles creates a different outcome. A combination counts selections, so the chosen group is the same however its members are listed.

Ask	If yes	If no
Would swapping two chosen items create a new result?	permutation	combination
Are positions, ranks, seats, roles, or code order specified?	permutation	check whether it is only a group
Is the outcome just a team, committee, or subset?	check for assigned roles	combination

From 8 people, choosing a 3-person committee is a combination: the group $\{A,B,C\}$ is one outcome. Filling president, secretary and treasurer is a permutation: $A,B,C$ in those roles differs from $B,A,C$ . These are separate counting situations even though both choose 3 of 8.

Words such as “arrange”, “code”, “first and last” or named offices usually signal order. Words such as “choose”, “select”, “team” or “committee” usually signal no order. The wording is a clue, but the swap test is the decisive check.

Do not choose a formula only because the question uses the word “choose”: a chosen group with assigned roles is ordered. Conversely, listing a committee in a different written order does not make a new committee.

For a non-negative integer $n$ , $n!=n(n-1)(n-2)\cdots2\cdot1$ . Also $0!=1$ : there is exactly one way to arrange or select nothing, and this value keeps factorial formulas valid when $r=n$ .

{}^nP_r=\frac{n!}{(n-r)!},\qquad {}^nC_r=\frac{n!}{r!(n-r)!},\qquad 0\le r\le n

The permutation formula keeps the $r$ descending choices $n(n-1)\cdots(n-r+1)$ . A selected set of $r$ objects appears in $r!$ different orders, so dividing ${}^nP_r$ by $r!$ gives ${}^nC_r$ . Therefore ${}^nP_r=r!\,{}^nC_r$ .

For 8 objects taken 3 at a time, ${}^8P_3=8 imes7 imes6=336$ , while ${}^8C_3=336/3!=56$ . In an algebraic equation, expand only the factors that do not cancel: if ${}^nP_5=6\,{}^{n-1}P_4$ , both sides contain $(n-1)(n-2)(n-3)(n-4)$ , leaving $n=6$ .

The upper and lower values must be whole numbers with $0\le r\le n$ . Cancel factorials before expanding them into large numbers, and distinguish the lowercase variable $n$ from a fixed numerical value.

Restrictions change which outcomes are allowed, not the meaning of permutation or combination. First decide whether the outcome is ordered; then organise the restriction with positions, cases, a block, or a complement.

Restriction	Reliable counting move
fixed first/last position, parity, or leading digit	fill the restricted position first, then multiply remaining choices
several mutually exclusive possibilities	count each complete case, then add
specified objects must stay together in a line	treat them as one block, then multiply by their internal orders
at least one selected object has a property	total selections minus selections with none

If 3 distinct men must stand together with 3 distinct women in a line, treat the men as one block. The block and 3 women form 4 units, arranged in $4!$ ways, while the men can be ordered internally in $3!$ ways. The count is $4! imes3!=144$ .

To select 7 letters from 4 vowels and 9 consonants with at least 2 vowels, the possible vowel counts are 2, 3 or 4. The total is ${}^4C_2{}^9C_5+{}^4C_3{}^9C_4+{}^4C_4{}^9C_3=1344$ . The cases are added because no selection belongs to two different vowel-count cases.

Stay inside the syllabus boundary: questions with repeated objects, circular arrangements, or a single problem requiring both permutations and combinations are excluded. For permitted problems, make cases disjoint and check that a leading zero or forbidden position has not been counted.

12. Series

In $(a+b)^n$ , each term chooses some factors to contribute $b$ and the rest to contribute $a$ . The coefficient ${}^nC_r$ counts the ways to choose the $r$ factors that contribute $b$ .

(a+b)^n=\sum_{r=0}^{n}{}^nC_r,a^{n-r}b^r

As $r$ increases from $0$ to $n$ , the power of $a$ decreases while the power of $b$ increases, and their exponents always add to $n$ . If $b$ is negative, its sign is raised to the power $r$ , so signs alternate when appropriate.

For $(2-3x)^4$ , the five terms are $2^4$ , ${}^4C_1 2^3(-3x)$ , ${}^4C_2 2^2(-3x)^2$ , ${}^4C_3 2(-3x)^3$ and $(-3x)^4$ . After simplifying, the expansion is $16-96x+216x^2-216x^3+81x^4$ .

Here $n$ is a positive integer, so the expansion terminates. The formula is supplied, but every coefficient and power must still be simplified; brackets around a negative term prevent sign errors.

The general term lets you find one required term without writing the whole expansion. In $(a+b)^n$ , choosing $r$ copies of $b$ gives term number $r+1$ , because the first term corresponds to $r=0$ .

T_{r+1}={}^nC_r,a^{n-r}b^r,\qquad 0\le r\le n

Substitute the full expressions for $a$ and $b$ , combine their powers of the variable, and solve the exponent condition. Use exponent $0$ for a term independent of the variable; use the requested exponent for a specified power. Only after finding an integer $r$ in the allowed range should you evaluate the coefficient.

In $\left(6/x^2+x^4/2 ight)^{12}$ , $T_{r+1}={} ^{12}C_r(6x^{-2})^{12-r}(x^4/2)^r$ . The power of $x$ is $-2(12-r)+4r=-24+6r$ . An independent term requires $-24+6r=0$ , so $r=4$ and the term is ${}^{12}C_4 6^8/2^4=51\,963\,120$ .

Do not confuse $r$ with the term number: $r=4$ identifies the fifth term. Greatest-term questions and general properties of binomial coefficients are outside this syllabus objective.

An arithmetic progression changes by a constant difference; a geometric progression changes by a constant multiplier. Test consecutive terms in the same way throughout the sequence.

Feature	Arithmetic progression (AP)	Geometric progression (GP)
constant relationship	$u_{k+1}-u_k=d$	$u_{k+1}/u_k=r$ when the ratio is defined
nth term	$a+(n-1)d$	$ar^{n-1}$
typical pattern	add or subtract the same amount	multiply by the same factor

The sequence $5,9,13,17,\ldots$ is arithmetic because every difference is $4$ . The sequence $12,6,3,3/2,\ldots$ is geometric because every ratio is $1/2$ . The sequence $2,4,7,11,\ldots$ is neither: its differences and ratios both change.

If three algebraic expressions are claimed to form a GP, use $T_2/T_1=T_3/T_2$ , or equivalently $T_2^2=T_1T_3$ when this avoids invalid division. For an AP, use $T_2-T_1=T_3-T_2$ . Then check any excluded zero denominators or parameter conditions.

A steadily increasing sequence is not automatically arithmetic, and a sequence containing powers is not automatically geometric. The invariant difference or ratio is the deciding evidence.

Finite progression problems reduce to the first term $a$ , the common difference $d$ or ratio $r$ , and the number of terms $n$ . Translate every given term or sum before solving for the unknown parameters.

Quantity	Arithmetic progression	Geometric progression
nth term	$u_n=a+(n-1)d$	$u_n=ar^{n-1}$
first $n$ terms	$S_n=\frac n2[2a+(n-1)d]$	$S_n=\frac{a(1-r^n)}{1-r}$ for $r\ne1$
known last term $l$	$S_n=\frac n2(a+l)$	use $l=ar^{n-1}$ if useful

If an AP has third term $10$ and $S_8=116$ , then $a+2d=10$ and $4(2a+7d)=116$ . Solving gives $a=4$ , $d=3$ . A run of 19 terms starting at the 12th ends at the 30th, so its sum is $S_{30}-S_{11}=1216$ .

If a GP has third term $4.5$ and sixth term $15.1875$ , then $ar^2=4.5$ and $ar^5=15.1875$ . Division gives $r^3=3.375$ , so $r=1.5$ and $a=2$ . A block of 10 terms starting at the 16th is $S_{25}-S_{15}$ , not $S_{10}$ .

The nth term is one term, whereas $S_n$ is a cumulative sum. If a GP has $r=1$ , every term is $a$ and $S_n=an$ . For a least-number-of-terms question, solve the inequality and then take the smallest valid positive integer, checking it in the original condition.

A geometric progression has a finite sum to infinity exactly when $|r|<1$ . Then successive terms shrink towards zero, so the remaining tail becomes arbitrarily small.

|r|<1\quad\Longrightarrow\quad S_\infty=\frac{a}{1-r}

From $S_n=a(1-r^n)/(1-r)$ , the condition $|r|<1$ makes $r^n o0$ , leaving $a/(1-r)$ . If $|r|\ge1$ , the terms do not tend to zero, so the partial sums cannot approach a finite limit. Always determine $r$ before using the formula.

For first terms $(2w-1/4),(2w-1/4)^2,(2w-1/4)^3$ , the ratio is $r=2w-1/4$ . Convergence requires $-1<2w-1/4<1$ , giving $-3/8<w<5/8$ . The endpoints are excluded because $|r|=1$ .

If a GP has first term $4$ and its first three terms sum to $7$ , then $4+4r+4r^2=7$ , so $r=1/2$ or $r=-3/2$ . Only $r=1/2$ converges; its sum to infinity is $4/(1-1/2)=8$ .

Check convergence separately for every possible ratio before evaluating $S_\infty$ . An arithmetic progression does not acquire a sum to infinity from these formulas, and a finite geometric sum may exist even when the infinite sum does not.

13. Vectors in two dimensions

A vector records both magnitude and direction. Different notations can describe the same vector, but the notation must show that the quantity is a vector rather than an ordinary number.

Form	What it identifies
$\mathbf{a}$ or $\mathbf{p}$	a named vector
$\overrightarrow{AB}$	the directed displacement from $A$ to $B$
$\begin{pmatrix}x\\y\end{pmatrix}$	horizontal and vertical components
$x\mathbf{i}+y\mathbf{j}$	the same components in unit-vector form

If $A=(2,-1)$ and $B=(7,3)$ , then $\overrightarrow{AB}=\begin{pmatrix}7-2\\3-(-1)\end{pmatrix}=\begin{pmatrix}5\\4\end{pmatrix}=5\mathbf{i}+4\mathbf{j}$ . These are three descriptions of the same directed displacement.

$\overrightarrow{AB}$ and $\overrightarrow{BA}$ point in opposite directions, so $\overrightarrow{BA}=-\overrightarrow{AB}$ . By contrast, $AB$ without an arrow normally denotes the scalar distance between the points. Do not drop bold type or the arrow when a vector is required.

A position vector locates a point from a fixed origin. If $R=(x,y)$ relative to origin $O$ , then $\overrightarrow{OR}=\begin{pmatrix}x\\y\end{pmatrix}$ . A unit vector keeps a direction but has magnitude $1$ .

|\mathbf{a}|=\sqrt{x^2+y^2},\qquad \widehat{\mathbf{a}}=\frac{\mathbf{a}}{|\mathbf{a}|}\quad(\mathbf{a}\ne\mathbf{0})

Find the magnitude first, then divide every component by that same positive magnitude. This scales the vector without changing its direction.

For $\mathbf{a}=\begin{pmatrix}5\\-12\end{pmatrix}$ , $|\mathbf{a}|=\sqrt{5^2+(-12)^2}=13$ . The unit vector in the same direction is $\frac1{13}\begin{pmatrix}5\\-12\end{pmatrix}=\begin{pmatrix}5/13\\-12/13\end{pmatrix}$ . Its magnitude checks as $1$ .

The zero vector has no direction, so it cannot be normalized. Dividing by a negative number would reverse the direction; the denominator in the unit-vector formula is the non-negative magnitude $|\mathbf a|$ .

Vector calculations are component-wise: combine horizontal components together and vertical components together. Equality of two vectors therefore gives one equation from each component.

Operation	Component rule
addition	$\begin{pmatrix}a\\b\end{pmatrix}+\begin{pmatrix}c\\d\end{pmatrix}=\begin{pmatrix}a+c\\b+d\end{pmatrix}$
subtraction	$\begin{pmatrix}a\\b\end{pmatrix}-\begin{pmatrix}c\\d\end{pmatrix}=\begin{pmatrix}a-c\\b-d\end{pmatrix}$
scalar multiplication	$k\begin{pmatrix}a\\b\end{pmatrix}=\begin{pmatrix}ka\\kb\end{pmatrix}$
magnitude	$\left\|\begin{pmatrix}a\\b\end{pmatrix}\right\|=\sqrt{a^2+b^2}$

For example, $\begin{pmatrix}4\\1\end{pmatrix}+k\begin{pmatrix}-2\\3\end{pmatrix}=r\begin{pmatrix}-10\\5\end{pmatrix}$ gives $4-2k=-10r$ and $1+3k=5r$ . Solving the two scalar equations gives $k=-3/2$ and $r=-7/10$ .

In vector geometry, build a route using directed segments. If $\overrightarrow{OA}=\mathbf a$ , $\overrightarrow{OB}=\mathbf b$ and $P$ divides $AB$ with $AP:PB=1:3$ , then $\overrightarrow{AB}=\mathbf b-\mathbf a$ and $\overrightarrow{OP}=\mathbf a+\frac14(\mathbf b-\mathbf a)=\frac34\mathbf a+\frac14\mathbf b$ .

To prove points are collinear, show that two directed vectors with a common point are scalar multiples. A positive scalar gives the same direction and a negative scalar gives the opposite direction, but both lie on the same straight line. Equal magnitudes alone do not prove parallelism or collinearity.

A velocity vector combines speed with direction. Resolve every velocity into components on the same axes; resultant velocities are then found by component-wise addition, and position changes linearly when velocity is constant.

\mathbf p(t)=\mathbf p_0+t\mathbf v

With $\mathbf i$ east and $\mathbf j$ north, a speed $s$ on bearing $\theta$ has velocity $s\sin\theta\,\mathbf i+s\cos\theta\,\mathbf j$ , because a bearing is measured clockwise from north. If an angle $\alpha$ is measured above the positive horizontal axis instead, use $s\cos\alpha\,\mathbf i+s\sin\alpha\,\mathbf j$ .

A cyclist travelling at $4\,\mathrm{m\,s^{-1}}$ on bearing $015^\circ$ has velocity $4\sin15^\circ\,\mathbf i+4\cos15^\circ\,\mathbf j$ . If another velocity acts at the same time, add its $\mathbf i$ and $\mathbf j$ components separately to obtain the resultant.

For two particles, write both position vectors using the same time variable. They collide only if both component equations give the same admissible time. For instance, if the $x$ -components meet at one time but the $y$ -components meet at another, the paths may cross geometrically but the particles do not collide.

Keep displacement, velocity and speed distinct: displacement and velocity are vectors, while speed is the scalar magnitude of velocity. Include the initial position $\mathbf p_0$ unless the particle starts at the origin.

Combinatorics, Series and Vectors

Exam analysis

Most tested topics

11. Permutations and combinations

Decide whether order matters

Build permutation and combination formulas from factorials

Count restricted arrangements and selections

12. Series

Expand a positive-integer binomial

Use the general term to target one binomial term

Tell arithmetic and geometric progressions apart

Model finite arithmetic and geometric progressions

Decide whether a geometric series converges

13. Vectors in two dimensions

Read and write vector notation precisely

Use position vectors and normalize direction

Calculate with vectors and prove geometry

Resolve velocity and model moving positions