Counting
Determine the number of possible outcomes without listing every outcome.
S4 · Topic 12 · Reasoning & Quant
Permutations, Combinations & Probability
Notes for UPSC Principal & Vice-Principal examinations only.
01Basics
The Foundation — Count Before You Calculate Probability
Separate arrangement from selection, define the sample space, and preserve restrictions from the first step.
AThe four central ideas
Permutation
An arrangement where order or position matters.
Combination
A selection where the chosen group matters but internal order does not.
Probability
A number from \(0\) to \(1\) measuring the chance of an event.
Does order matter?Yes → permutation
Only selection matters?Yes → combination
Chance requested?Count favourable and total outcomes
BFundamental counting principles
Multiplication principle — AND
If a process has successive stages with \(m_1,m_2,\ldots,m_k\) choices:
\[N=m_1m_2\cdots m_k\]
Use when every complete outcome requires one choice from each stage.
Addition principle — OR
If mutually exclusive alternatives have \(m_1,m_2,\ldots,m_k\) outcomes:
\[N=m_1+m_2+\cdots+m_k\]
If alternatives overlap, subtract the overlap.
Inclusion–exclusion for two sets
\[|A\cup B|=|A|+|B|-|A\cap B|\]
Adding without subtracting intersection double-counts shared outcomes.
CFactorials: the counting engine
Definition\[n!=n(n-1)(n-2)\cdots3\cdot2\cdot1\]
- \(0!=1\) by definition.
- \(1!=1\).
- \(n!=n(n-1)!\).
- Cancel factorials before expanding: \(\frac{n!}{(n-r)!}=n(n-1)\cdots(n-r+1)\).
DPermutations: order matters
| Situation | Formula | Interpretation |
|---|---|---|
| Arrange all \(n\) distinct objects | \(n!\) | \(n\) choices, then \(n-1\), continuing to \(1\). |
| Arrange \(r\) from \(n\) distinct objects | \({}^{n}P_r=\frac{n!}{(n-r)!}\) | Selection and order occur together. |
| Repetition allowed, \(r\) positions | \(n^r\) | Each position has \(n\) independent choices. |
| Identical objects: groups \(p,q,\ldots\) | \(\frac{n!}{p!q!\cdots}\) | Divide arrangements indistinguishable within each identical group. |
| Circular arrangement, rotations identical | \((n-1)!\) | Fix one object to remove rotational duplication. |
| Necklace, reflection also identical | \(\frac{(n-1)!}{2}\) | Valid for distinct beads with \(n\gt2\). |
Distinct arrangement
Arrange \(5\) different files on a shelf.
\[5!=120\]
There are \(120\) orders.
Choose and rank
Select a president, secretary and treasurer from \(8\) people.
\[{}^{8}P_3=8\times7\times6=336\]
Posts are distinct, so order matters.
Repeated letters
Distinct arrangements of LEVEL:
\[\frac{5!}{2!2!}=30\]
L and E each repeat twice.
ECombinations: selection without order
Core formula
\[{}^{n}C_r=\binom nr=\frac{n!}{r!(n-r)!}\]
Symmetry
\[\binom nr=\binom n{n-r}\]
Selecting \(r\) included objects is equivalent to selecting \(n-r\) excluded objects.
Permutation–combination link
\[{}^{n}P_r=\binom nr r!\]
First select the \(r\) objects, then arrange them.
Pascal identity
\[\binom nr=\binom{n-1}{r}+\binom{n-1}{r-1}\]
Sum of all selections
\[\sum_{r=0}^{n}\binom nr=2^n\]
Number of all subsets of an \(n\)-element set.
Non-empty subsets
\[2^n-1\]
Remove the empty set.
FSelection conditions
| Wording | Counting translation | Preferred strategy |
|---|---|---|
| Exactly \(k\) from a group | Choose \(k\) from that group and the remainder from others. | Multiply combination factors. |
| At least \(k\) | Sum cases \(k,k+1,\ldots\) | Or use complement when excluded cases are fewer. |
| At most \(k\) | Sum cases \(0,1,\ldots,k\) | Check whether zero is allowed. |
| Particular person must be included | Fix that person; select remaining members. | \(\binom{n-1}{r-1}\) |
| Particular person excluded | Remove that person from the pool. | \(\binom{n-1}{r}\) |
| Two persons not together in a committee | Total committees minus committees containing both. | \(\binom nr-\binom{n-2}{r-2}\) |
GProbability foundations
ExperimentA repeatable random process
OutcomeOne possible result
Sample space \(S\)Set of all outcomes
Event \(A\)A subset of \(S\)
Classical probability
\[P(A)=\frac{n(A)}{n(S)}\]
Requires finite, mutually exclusive and equally likely elementary outcomes.
Range and certainty
\[0\le P(A)\le1,\quad P(S)=1,\quad P(\varnothing)=0\]
Complement
\[P(A^c)=1-P(A)\]
Most useful for “at least one.”
HAddition and multiplication laws
General addition law
\[P(A\cup B)=P(A)+P(B)-P(A\cap B)\]
If events are mutually exclusive, \(P(A\cap B)=0\).
General multiplication law
\[P(A\cap B)=P(A)P(B\mid A)\]
If independent, \(P(B\mid A)=P(B)\), so \(P(A\cap B)=P(A)P(B)\).
Mutually exclusive is not the same as independent
Mutually exclusive non-impossible events cannot occur together and are therefore dependent. Independent events may occur together, but one does not change the probability of the other.
IStandard sample spaces
| Experiment | Total equally likely outcomes | Key structure |
|---|---|---|
| One fair coin | \(2\) | Head, tail. |
| \(n\) fair coin tosses | \(2^n\) | Exactly \(r\) heads occur in \(\binom nr\) sequences. |
| One fair die | \(6\) | Faces \(1\) through \(6\). |
| Two distinguishable dice | \(36\) | Ordered pairs; sum frequencies are \(1,2,3,4,5,6,5,4,3,2,1\). |
| Standard card deck | \(52\) | \(4\) suits, \(13\) ranks, \(26\) red, \(26\) black. |
| Face cards | \(12\) | Jack, queen, king in each suit. |
| Aces / kings / queens | \(4\) of each rank | One in each suit. |
02Expert
The Deep Dive — Restrictions, Conditional Models and Bayes
Control adjacency, repetition, circular order, case splits, dependent draws and reverse probability.
ARestricted linear arrangements
Objects together — block method
Treat \(k\) specified objects as one super-object, then arrange internally.
\[(n-k+1)!\,k!\]
For \(n\) distinct objects in a line.
Specified objects not together
\[\text{Required}=\text{Total}-\text{Together}\]
Complement is usually faster than direct casework.
No two alike adjacent — gap method
Arrange the larger/separating group first. If it creates \(g\) gaps, choose and fill gaps with the other group.
\[g=m+1\]
Two people together
Arrange \(7\) people in a row if A and B stay together.
\[6!\times2!=1440\]
The pair is one block with \(2\) internal orders.
Two people separated
Arrange \(7\) people if A and B are not adjacent.
\[7!-6!\times2!=3600\]
Subtract the together cases.
No adjacent vowels
Arrange \(5\) distinct consonants and \(3\) distinct vowels with no adjacent vowels.
\[5!\binom63 3!\]
Consonants form \(6\) gaps.
BPosition, digit and divisibility restrictions
Leading zero
Zero cannot occupy the first position of a numeral. Choose the first digit separately, then fill remaining places.
\[(\text{non-zero first choices})\times(\text{remaining choices})\]
Even / odd numeral
Choose the units digit first from allowed even or odd digits, then complete other positions.
Divisibility conditions
- By \(3\) or \(9\): digit sum.
- By \(4\): last two digits.
- By \(5\): last digit \(0\) or \(5\).
- By \(11\): alternating-sum test.
CCircular arrangements with conditions
Ordinary round table
\[(n-1)!\]
Fix one person; clockwise and anticlockwise orders remain distinct.
Two specified together
Treat the pair as a block among \(n-1\) circular entities, then arrange the pair.
\[(n-2)!\times2!\]
Alternating two equal groups
Fix and arrange one group around the circle, then place the second group in the gaps.
\[(m-1)!m!\]
For \(m\) distinct members in each group.
DSelection with repetition: stars and bars
Non-negative integer solutions
Number of solutions to \(x_1+x_2+\cdots+x_n=r\), where every \(x_i\ge0\):
\[\binom{n+r-1}{r}=\binom{n+r-1}{n-1}\]
Equivalent to selecting \(r\) objects from \(n\) types with unlimited repetition.
Positive integer solutions
If every \(x_i\ge1\), allocate one to each variable first:
\[\binom{r-1}{n-1}\]
Valid when \(r\ge n\).
EDerangements and inclusion–exclusion
Derangement
A permutation in which no object remains in its original position.
\[!n=n!\sum_{k=0}^{n}\frac{(-1)^k}{k!}\]
Nearest integer to \(\frac{n!}{e}\). Small values: \(!1=0,!2=1,!3=2,!4=9,!5=44\).
Three-set inclusion–exclusion
\[|A\cup B\cup C|=|A|+|B|+|C|-|A\cap B|-|B\cap C|-|C\cap A|+|A\cap B\cap C|\]
FCombinatorial geometry and grouping
Handshakes / pairs
\[\binom n2=\frac{n(n-1)}2\]
Each unordered pair contributes one handshake.
Triangles from points
\[\binom n3\]
Valid if no three selected points are collinear; subtract collinear triples otherwise.
Polygon diagonals
\[\binom n2-n=\frac{n(n-3)}2\]
All vertex pairs minus \(n\) sides.
Divide into unlabeled pairs
\[\frac{(2n)!}{2^n n!}\]
Internal order within each pair and order of pairs do not matter.
Equal unlabeled groups
Divide \(mn\) distinct objects into \(m\) unlabeled groups of size \(n\):
\[\frac{(mn)!}{(n!)^m m!}\]
Choose then arrange
\[\binom nr r!={}^{n}P_r\]
Useful when a question separates selection and assignment.
GConditional probability and independence
Conditional probability
\[P(A\mid B)=\frac{P(A\cap B)}{P(B)},\quad P(B)\gt0\]
Once B is known, B becomes the reduced sample space.
Independence test
\[P(A\cap B)=P(A)P(B)\]
Equivalently, \(P(A\mid B)=P(A)\) when defined.
Chain rule
\[P(A_1\cap\cdots\cap A_n)=P(A_1)P(A_2\mid A_1)\cdots P(A_n\mid A_1\cap\cdots\cap A_{n-1})\]
HSampling with and without replacement
With replacement
Population composition resets after every draw. Draws are independent when selection probabilities remain unchanged.
\[P(A\text{ then }B)=P(A)P(B)\]
Without replacement
Population size and composition change; draws are dependent.
\[P(A\text{ then }B)=P(A)P(B\mid A)\]
Two red objects without replacement
A bag contains \(5\) red and \(3\) blue objects. Probability that two drawn are red:
\[\frac58\cdot\frac47=\frac5{14}\quad\text{or}\quad\frac{\binom52}{\binom82}=\frac5{14}\]
Sequential and combination methods agree.
ITotal probability and Bayes’ theorem
Total probability
If \(B_1,\ldots,B_n\) partition the sample space:
\[P(A)=\sum_{i=1}^{n}P(B_i)P(A\mid B_i)\]
Bayes’ theorem
\[P(B_j\mid A)=\frac{P(B_j)P(A\mid B_j)}{\sum_{i=1}^{n}P(B_i)P(A\mid B_i)}\]
Posterior probability is proportional to prior probability times likelihood.
JBinomial probability and repeated trials
Exactly \(r\) successes
For \(n\) independent identical trials, success probability \(p\), failure probability \(q=1-p\):
\[P(X=r)=\binom nr p^r q^{n-r}\]
At least one success
\[P(X\ge1)=1-P(X=0)=1-q^n\]
Complement avoids a long case sum.
Mean and variance
\[E(X)=np,\qquad \operatorname{Var}(X)=npq\]
Standard deviation \(=\sqrt{npq}\).
KOdds and expected value
Odds in favour
If odds in favour are \(a:b\):
\[P(A)=\frac{a}{a+b}\]
Odds against
If odds against are \(a:b\):
\[P(A)=\frac{b}{a+b}\]
Expected value
\[E(X)=\sum_i x_iP(X=x_i)\]
A long-run average, not a guaranteed single-trial outcome.
03Accuracy
The UPSC Trap / Watch Out
Prevent overcounting, wrong denominators and confusion between independence and exclusivity.
AClassic distractors and common errors
Permutation used for a committee
If posts are not distinct, internal order does not matter; use combinations.
Combination used for offices
President and secretary are different roles, so assignments are ordered.
Repetition assumed silently
Check whether an object or digit can be reused.
Leading zero accepted
A multi-digit numeral cannot begin with zero.
Identical objects treated distinct
Divide by factorials of repeated groups.
Linear formula used in a circle
Rotations are identical at a round table; use \((n-1)!\).
Necklace confused with round table
In a necklace, reflections are also identical; divide by \(2\) when applicable.
Together block not arranged internally
After forming a block, multiply by arrangements inside the block.
Gap count wrong
\(m\) arranged separators create \(m+1\) linear gaps but only \(m\) circular gaps.
“At least” read as “exactly”
Sum all allowed higher cases or use the complement.
Overlapping alternatives added
Use inclusion–exclusion when OR cases overlap.
Outcomes not equally likely
Do not apply favourable/total counting unless elementary outcomes have equal probability.
Two-dice sums treated equally likely
Sums \(2\) through \(12\) do not have equal frequencies.
Mutually exclusive called independent
Non-impossible disjoint events are dependent.
Replacement condition ignored
Without replacement changes the next denominator and numerator.
Conditional denominator reversed
In \(P(A\mid B)\), the conditioning event B is the denominator.
At least one counted directly
Use \(1-P(\text{none})\) unless direct cases are genuinely shorter.
Bayes numerator not normalised
Divide prior × likelihood by total evidence probability.
BHigh-speed shortcuts
Order test
Swap two selected objects mentally. If the outcome changes, use permutation.
Combination symmetry
Replace \(\binom nr\) by \(\binom n{n-r}\) when the lower number becomes smaller.
Factorial cancellation
Expand only the factors needed; never calculate large factorials in full.
At least one
Use complement: \(1-P(0)\).
Not together
Total minus together is usually the shortest route.
Gap method
Arrange separators first, mark gaps, then choose gaps for restricted objects.
Pairs immediately
Handshakes, lines and unordered pairs suggest \(\binom n2\).
Sequential vs combination
Without-replacement draws can often be solved either by multiplying fractions or counting subsets.
Probability direction check
Conditioning on favourable information should move probability sensibly; every result must stay between \(0\) and \(1\).
Fast accuracy rule
Define one complete outcome, decide whether order and repetition matter, encode every restriction before counting, and use the same outcome model in both numerator and denominator.
04Revise
Last-Minute Master Sheet
Compact relationships for rapid revision.
Factorial\[n!=n(n-1)!,\qquad0!=1\]
Permutation\[{}^{n}P_r=\frac{n!}{(n-r)!}\]
Combination\[\binom nr=\frac{n!}{r!(n-r)!}\]
Relation\[{}^{n}P_r=\binom nr r!\]
Repeated objects\[\frac{n!}{p!q!\cdots}\]
Circular order\[(n-1)!\]
Selections with repetition\[\binom{n+r-1}{r}\]
Classical probability\[P(A)=\frac{n(A)}{n(S)}\]
Complement\[P(A^c)=1-P(A)\]
Addition law\[P(A\cup B)=P(A)+P(B)-P(A\cap B)\]
Conditional probability\[P(A\mid B)=\frac{P(A\cap B)}{P(B)}\]
Binomial\[P(X=r)=\binom nr p^r(1-p)^{n-r}\]
One-line memory rule
Permutation orders a selection; combination ignores that order; probability compares valid favourable outcomes with the same valid universe.