Permutation and Combination | NEB Class 12 Mathematics Notes

Chapter Overview

This chapter is about counting without listing. Two big ideas run through everything: Permutation (order matters — ABC and BAC are different) and Combination (order doesn't matter — picking A,B,C is the same as picking C,A,B). Spot which one applies first, and the calculation becomes easy and fast.

What this chapter covers

Multiplication & Addition PrinciplesFactorial NotationLinear PermutationIdentical ObjectsCircular PermutationPermutation with RepetitionCombinationsComplementary CombinationPascal's RuleTogether & Never Together TricksAlternating / Gaps Trick

Learning Objectives

1

Tell apart situations where order matters (permutation) vs doesn't matter (combination) by asking one simple question

2

Use the multiplication principle and addition principle to break big counting problems into manageable steps

3

Apply formulas for linear permutation, identical-object permutation, and circular permutation correctly

4

Solve combination problems including committee selection and 'at least / at most' type questions

5

Recognise keyword patterns in questions (rank, arrange, choose, committee, necklace, PIN...) and match them to the right formula

6

Handle special constraints like 'must sit together', 'never together', and 'must alternate' using the bundling and gaps tricks

The One Question That Solves Everything

Key Insight

Before touching any formula, ask yourself: "If I swap two of the chosen items, do I get a different result?"

If YES → the order matters → use Permutation.
If NO → the order doesn't matter → use Combination.

Example: Arranging 3 students in a line — swapping gives a different line → Permutation. Picking a committee of 3 — swapping gives the same group → Combination.

Key Definitions

Permutation

An arrangement of items where order matters. ABC and BAC are two different permutations of the same three letters. Think: positions are labelled.

Combination

A selection of items where order does not matter. {A, B, C} and {C, A, B} are the same combination. Think: we only care who's in the group, not who stands where.

Factorial (n!)

The product of all positive integers from 1 up to n. It counts the total ways to arrange n distinct objects in a line.

$$n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1, \qquad 0! = 1$$

Multiplication Principle

If task A can be done in m ways AND task B in n ways (independently), then both together can be done in m × n ways. Use this when steps happen one after another.

Addition Principle

If task A can be done in m ways OR task B in n ways (mutually exclusive — not both at once), then the total is m + n ways. Keywords: OR, either, at least, choose one type.

Mutually Exclusive

Two events that cannot happen at the same time. When events are mutually exclusive, you add their counts. Example: picking a red ball or a blue ball from a bag.

Permutation vs Combination — Side by Side

The most important distinction in this entire chapter. Study this table until the difference feels natural.

Permutation P(n, r)

$$P(n,r) = \dfrac{n!}{(n-r)!}$$

Order matters — ABC ≠ BAC ≠ CAB

Same items in different order = different result

Keywords: arrange, rank, order, sequence, line up, position

Examples: seating arrangement, forming a queue, choosing officers (President, VP, Sec)

Always ≥ C(n,r) because every combination produces r! permutations

VS

Combination C(n, r)

$$C(n,r) = \dfrac{n!}{r!(n-r)!}$$

Order doesn't matter — {A,B,C} = {C,A,B}

Same items in any order = same result

Keywords: choose, select, pick, committee, team, group, handshake

Examples: forming a committee, selecting players for a team, drawing cards

Always ≤ P(n,r) — it's permutation divided by r!

Memory trick: Permutation → Positions are labelled. Combination → Crowd, no labels.

Formula Sheet — All Formulas with Full Derivations

Linear Permutation

P(n,r) = \frac{n!}{(n-r)!}

Arranging r items chosen from n distinct items. Order matters, no repetition.

1Picture r empty slots in a row. You have n objects to fill them with, and once an object is placed it cannot be reused.

Slot 1 has n choices. Slot 2 has (n−1) choices (one object is already used). Slot 3 has (n−2), and so on until the r-th slot has (n−r+1) choices.

2Multiply all these independent choices together using the Multiplication Principle:
$$n \times (n-1) \times (n-2) \times \cdots \times (n-r+1)$$
This is a partial factorial — it starts at n and has exactly r terms.

3To write this cleanly using factorial notation, notice that $n! = n(n-1)\cdots(n-r+1)(n-r)!$. So dividing both sides by $(n-r)!$ isolates our r terms:
$$P(n,r) = \frac{n!}{(n-r)!}$$
The division by $(n-r)!$ simply cancels out all the terms we didn't need.

Permutation of Identical Objects

\frac{n!}{p! \; q! \; r! \; \cdots}

Arranging all n items when some are identical. Divide by factorial of each repeated group.

1Start by pretending all n objects are completely distinct — even the ones that look the same. With everything labelled uniquely, the total arrangements would be n!.

2Now suppose p of the objects are actually identical (say, p copies of the letter 'A'). If you secretly swap any two of these p copies, the word looks exactly the same to a reader — but your count treated it as a new arrangement. Every real arrangement is being counted p! times (once for each way the p identical items can be shuffled amongst themselves).

3Fix the overcount by dividing by p!. If there are more groups — q identical items of another type, r of another, etc. — divide by each of their factorials too:
$$\frac{n!}{p! \times q! \times r! \times \cdots}$$
Each division removes the overcounting caused by one group of identical items.

Circular Permutation

$$(n-1)!$$

Arranging n distinct items around a fixed circle (like a round table that cannot be flipped).

1In a straight line, n distinct people have n! arrangements — every seat has a unique position number, so every shuffle is counted separately.

2Now bend that line into a circle. Suddenly, rotating everyone one seat clockwise gives an arrangement that looks exactly the same (same neighbours, same left-right order). In fact, any one of the n rotations of an arrangement produces an identical-looking circle.

3So each unique circular arrangement is being counted n times inside our n! total. Divide out this redundancy:
$$\frac{n!}{n} = (n-1)!$$
Intuitively: fix one person's seat as a reference point (they don't rotate). Now arrange the remaining (n−1) people in the remaining (n−1) seats → $(n-1)!$

Circular — Flippable (Necklace / Bracelet)

\frac{1}{2}(n-1)!

Use when the circular object can be physically flipped over — necklaces, bracelets, garlands.

1Start from the circular permutation formula: $(n-1)!$ arrangements for a fixed circle (like people around a table bolted to the floor).

2A bracelet or necklace is different — you can pick it up and flip it over in 3D space. When you flip it, the clockwise arrangement A→B→C→D becomes the anticlockwise arrangement A→D→C→B. These two look different on a table but are physically identical as a bracelet you can wear.

3This means every bracelet arrangement is being counted twice in our $(n-1)!$ (once for each direction). Divide by 2 to remove this double-counting:
$$\frac{(n-1)!}{2}$$
Key memory: Round table → $(n-1)!$ Bracelet/necklace → $\frac{1}{2}(n-1)!$

Permutation with Repetition

$$n^r$$

Filling r positions from n options when repetition is allowed (PINs, passwords, codes, mailboxes).

1You have r positions to fill, and for each position you have n choices. Unlike the standard permutation, nothing is removed after you pick — the same item can appear again in any later slot. Think of a 4-digit PIN: each digit can be 0–9 regardless of what came before.

2Since each of the r positions is completely independent of the others (all n choices available every time), apply the Multiplication Principle across all r positions:
$$\underbrace{n \times n \times n \times \cdots \times n}_{r \text{ times}} = n^r$$

3This is simply exponentiation — n choices repeated r independent times. Compare to $P(n,r) = n(n-1)(n-2)\cdots$ where choices shrink because repetition is forbidden. Here they don't shrink at all.

Combination

C(n,r) = \frac{n!}{r!(n-r)!}

Selecting a group of r from n items where order does not matter.

1Start with $P(n,r)$ — the number of ways to choose and arrange r items from n. Every unique group is already accounted for in $P(n,r)$, but each group appears r! times (once for each internal ordering of its r members).

2Since combinations don't care about internal order, all those r! orderings of the same group should count as just one combination. We've overcounted every group by exactly a factor of r!.

3Divide $P(n,r)$ by r! to collapse all orderings of each group into a single count:
$$C(n,r) = \frac{P(n,r)}{r!} = \frac{n!}{r!(n-r)!}$$
This is the fundamental relationship: every C(n,r) combinations, when arranged internally, produce $C(n,r) \times r! = P(n,r)$ ordered arrangements.

Complementary Combination

C(n,r) = C(n,\, n-r)

Powerful shortcut when r is large (e.g. C(20,17) = C(20,3) — much easier to compute).

1Imagine you have n distinct items and you want to choose r of them to keep. Every valid selection of r items to keep automatically and uniquely defines the remaining (n−r) items as the ones left out.

2Choosing 'which r to keep' and choosing 'which (n−r) to reject' are two descriptions of the exact same decision. They must produce the same count of outcomes.

3Therefore $C(n,r) = C(n,n-r)$. Practical use: if r > n/2, computing $C(n,n-r)$ involves smaller numbers in the denominator — much faster. For example, $C(20,17) = C(20,3) = \frac{20 \times 19 \times 18}{3!} = 1140$ instead of the messy $\frac{20!}{17! \times 3!}$.

Pascal's Rule

C(n,\,r-1) + C(n,\,r) = C(n+1,\,r)

Used in algebraic proofs, the binomial theorem, and Pascal's Triangle.

1Suppose you want to select a committee of r people from a total pool of (n+1) people. The number of ways to do this is $C(n+1, r)$ — that's the right-hand side we want to derive.

2Now pick out one specific person from the pool — call them Person X. Every possible committee of r either includes Person X or it doesn't. These two cases cover every possibility with zero overlap, so we can add them.

Case 1 — Person X is included: The remaining (r−1) seats must be filled from the other n people → $C(n,\, r-1)$ ways.
Case 2 — Person X is excluded: All r seats must be filled from the other n people → $C(n,\, r)$ ways.

3Adding both mutually exclusive cases gives the total — which must equal $C(n+1, r)$:
$$C(n,r-1) + C(n,r) = C(n+1,r)$$
This also explains Pascal's Triangle: each entry is the sum of the two entries directly above it.

Which Formula Should I Use?

When you read a question, walk through these decision points in order. Each branch leads to the correct formula.

Is repetition allowed? (codes, PINs, passwords, mailboxes)

Yes

Use n^r — each of the r positions independently has n choices

No

Continue to next question ↓

Are items arranged in a CIRCLE? (round table, necklace, garland)

Yes

Round table → (n−1)! | Necklace/bracelet → ½(n−1)!

No

Continue to next question ↓

Does ORDER matter? (arranging, ranking, lining up, positions are labelled)

Yes

Use Permutation → P(n,r) = n! / (n−r)!

No

Use Combination → C(n,r) = n! / r!(n−r)!

Are there IDENTICAL / REPEATED items in the arrangement?

Yes

Divide n! by the factorial of each repeated group: n! / (p! × q! × ...)

No

Use the standard P or C formula from above

The 'Must Stay Together' and 'Never Together' Tricks

Trick: Items forced together → bundle them. Items that must stay apart → subtract the together-count from the total.

Steps to Remember:

Together: Treat the 'must stay together' group as one single block. Now you have fewer total units to arrange. After arranging the units, multiply by the number of ways to internally arrange the members of the block.
Never Together: Total arrangements (no restriction) − Arrangements where the forbidden items ARE together (calculated using the bundle method above).
Example — Together: 5 people, A and B must be adjacent. Bundle AB as 1 unit → 4 units total. Circular: $(4-1)! = 6$. Internal swaps of A,B: $2! = 2$. Total: $6 \times 2 = 12$.
Example — Never Together: Total circular arrangements of 5 = $(5-1)! = 24$. Minus arrangements where A,B are together = 12. Answer: $24 - 12 = 12$.

Mnemonic: Together = bundle into one block, then multiply by internal arrangements. Never Together = Total − Together.

The 'No Two Alike Adjacent' / Alternating / Gaps Trick

Trick: Place the larger group first to build gaps. Then drop the smaller group into those gaps — one per gap.

Steps to Remember:

Step 1: Arrange the larger group in a line first. This creates gaps: one before the first person, one between each pair, and one after the last. For n people in a line, this gives (n+1) gaps.
Step 2: Place members of the smaller group one per gap using permutation. Since gaps are distinct positions, use $P(\text{gaps}, \text{smaller group size})$.
Step 3: Multiply both counts using the Multiplication Principle.
Example: 5 boys and 4 girls, no two girls adjacent. Arrange 5 boys: $5! = 120$. Gaps created: 6 (_B_B_B_B_B_). Place 4 girls in 6 gaps: $P(6,4) = 360$. Total: $120 \times 360 = 43,200$.
Circular version: Fix the larger group in a circle first → $(n-1)!$ ways. Now there are exactly n gaps between them (equal to group size). Place smaller group in n gaps → $n!$ ways. Multiply.

Mnemonic: Bigger group first builds the gaps. Smaller group fills the gaps. Gaps in line = n+1. Gaps in circle = n.

Common Mistakes

These mistakes account for the most lost marks in NEB exams for this chapter. Read each one carefully.

P vs C confusion — Always ask: 'does swapping two of the chosen items change the outcome?' Yes → P. No → C. Don't guess based on the feel of the question.
Forgetting ÷2 for necklaces and bracelets — A round table cannot be flipped, so use $(n-1)!$. A necklace or bracelet CAN be flipped, so use $\frac{1}{2}(n-1)!$. Garlands also follow the bracelet rule.
Forgetting internal arrangements inside a bundle — When items are grouped together, always multiply by the factorial of the bundle size. If A and B are bundled, multiply by $2! = 2$ for their internal swap.
'At least / at most' handled wrong — These must be split into separate mutually exclusive cases and added together. Never try to handle 'at least 2' with one single formula.
0! = 1, not 0 — This trips up many students. $0!$ is defined as 1 by convention. Without this, $P(n,n) = n!/0!$ would break.
Digit problems where 0 cannot be the leading digit — Always handle the first position separately. The first digit of a number cannot be 0 or the number has fewer digits than expected.

NEB Exam Traps — These Appear Every Year

NEB Exam Warning

Board examiners deliberately test these edge cases. Students who don't know them lose 1–3 marks per question.

Bracelet/necklace question asks for 'arrangements' — students use $(n-1)!$ and forget the ÷2. Always confirm: can this object be flipped?
Committee question with gender restrictions — students add instead of multiply. Use multiplication principle: (men choices) × (women choices).
'At least one' type — students subtract from total. This works, but only if you subtract the correct complement. Wrong complement → wrong answer.
Identical letters word problem — students forget to divide by the factorial of EACH repeated letter, not just the most frequent one. ENGINEERING has E×3, N×3, G×2, I×2 — all four must be divided.
Circular arrangement where one person is 'fixed' — some students still divide by n. If one person is already fixed, treat it as a linear arrangement of the remaining (n−1) people.
Pascal's Rule proof — if asked for 'combinatorial proof', use the committee argument (include/exclude one person). If asked for 'algebraic proof', use factorials and find the common denominator.

Question Recognition Guide

Keyword / Phrase in Question	Use This
Arrange / line up / rank / order people or letters	Linear Permutation → P(n,r)
Word has repeated letters (APPLE, ENGINEERING, MATHEMATICS)	Identical Objects → n! ÷ (p! × q! × ...)
Round table / circular seating (not flippable)	Circular Permutation → (n−1)!
Necklace / bracelet / garland (can be flipped)	Flippable Circle → ½(n−1)!
Codes / PINs / passwords / passwords / mailboxes — repetition allowed	Repetition → n^r
Committee / team / group / handshake / draw cards — no positions	Combination → C(n,r)
'must sit / stand together' constraint	Bundle Trick → treat group as 1 block
'never together' / 'must not be adjacent'	Total − Together (using bundle trick for 'together')
'no two alike sit adjacent' / 'alternate' / 'boys and girls alternating'	Gaps Trick → seat larger group first, fill gaps
'at least' / 'at most' / 'exactly k'	Split into cases, add using Addition Principle
Diagonals of a polygon / handshakes in a group	C(n,2) − n (for diagonals) or C(n,2) (for handshakes)
C(n, x) = C(n, y) and x ≠ y	Complementary rule: n = x + y
C(n, r−1) + C(n, r) — asked to simplify	Pascal's Rule → C(n+1, r)

Why Circular Arrangements ÷ n?

Below are 5 people (A, B, C, D, E) seated in a circle. If everyone rotates one seat clockwise, the circle looks exactly the same — same neighbours, same relative positions. There are n=5 such rotations, each producing an identical-looking arrangement. That's why we divide n! by n to get (n−1)!

Both circles show the same relative arrangement (A next to B and E, B next to A and C, etc.) — rotating everyone one seat doesn't change anything. 5 such rotations exist, so we divide 5! by 5 = 4!

Solved Examples (Easy → Hard)

1

In how many ways can the letters of the word 'APPLE' be arranged?

Easy

Solution

First, count the total letters: APPLE has 5 letters.
Now identify any repeats: the letter 'P' appears 2 times. No other letter repeats.

If all 5 letters were distinct, arrangements = $5! = 120$.
But the two P's are identical — swapping them gives the same word. Each true arrangement is being counted $2! = 2$ times.

Fix by dividing:
$$\frac{5!}{2!} = \frac{120}{2} = 60$$

∴ 60 arrangements

2

5 friends sit around a circular table. A and B must always sit next to each other. How many arrangements are possible?

Medium

Solution

Since A and B must always be adjacent, bundle them into one single unit (AB).

Now we have 4 units to arrange around the circular table: the (AB) bundle + friends C, D, E.

Circular arrangements of 4 units:
$$(4-1)! = 3! = 6 \text{ ways}$$

Inside the bundle, A and B can swap places (AB or BA):
$$2! = 2 \text{ ways}$$

By the Multiplication Principle:
$$\text{Total} = 6 \times 2 = 12$$

∴ 12 arrangements

3

How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5 without repetition?

Medium

Solution

For the number to be even, the units digit must be even: either 2 or 4 from our set.
→ 2 choices for the units digit.

After fixing the units digit, 4 digits remain to fill the hundreds and tens places (2 slots, no repetition):
$$P(4,2) = 4 \times 3 = 12 \text{ ways}$$

By the Multiplication Principle:
$$\text{Total} = 2 \times 12 = 24$$

∴ 24 even numbers

4

A committee of 5 must be selected from 6 teachers and 8 students such that it includes at least 2 teachers. In how many ways?

Hard

Solution

'At least 2 teachers' means 2, 3, 4, or 5 teachers. Split into 4 mutually exclusive cases:

Case 1: Exactly 2 teachers, 3 students
$C(6,2) \times C(8,3) = 15 \times 56 = 840$

Case 2: Exactly 3 teachers, 2 students
$C(6,3) \times C(8,2) = 20 \times 28 = 560$

Case 3: Exactly 4 teachers, 1 student
$C(6,4) \times C(8,1) = 15 \times 8 = 120$

Case 4: Exactly 5 teachers, 0 students
$C(6,5) \times C(8,0) = 6 \times 1 = 6$

Since these cases cannot overlap, add all:
$$840 + 560 + 120 + 6 = 1526$$

∴ 1526 ways

5

In how many ways can the letters of the word 'ENGINEERING' be arranged?

Hard

Solution

Total letters in ENGINEERING = 11.

Count each letter's frequency:
• E appears 3 times
• N appears 3 times
• G appears 2 times
• I appears 2 times
• R, C, each appear 1 time

Using the identical objects formula:
$$\frac{11!}{3! \times 3! \times 2! \times 2!} = \frac{39{,}916{,}800}{6 \times 6 \times 2 \times 2} = \frac{39{,}916{,}800}{144}$$

∴ 277,200 arrangements

MCQ Practice

Mixed Level — MCQ

1

A room has 5 doors. In how many ways can a person enter through one door and exit through a different door?

Correct!

Incorrect. Correct: 20

Solution

Break the process into 2 sequential steps.

Step 1 — Enter: Any of the 5 doors → 5 choices.
Step 2 — Exit: Must use a different door → only 4 choices remain.

Since both steps happen together (AND), multiply:
$$5 \times 4 = 20$$

∴ 20

2

If C(n, 12) = C(n, 8), what is the value of n?

Correct!

Incorrect. Correct: 20

Solution

The complementary combination property states: if $C(n,r) = C(n,k)$ and $r \neq k$, then $n = r + k$.

(Because choosing r items to keep is the same as choosing (n−r) items to reject. So $k = n - r$, giving $n = r + k$.)

Here $C(n,12) = C(n,8)$ with $12 \neq 8$, so:
$$n = 12 + 8 = 20$$

∴ 20

3

In how many ways can 6 different coloured beads form a bracelet?

Correct!

Incorrect. Correct: 60

Solution

Step 1: Fix one bead's position (circular, so fix to remove rotation symmetry): $(6-1)! = 5! = 120$.

Step 2: A bracelet can be flipped over — clockwise and anticlockwise versions are the same physical bracelet. This doubles the count, so divide by 2:
$$\frac{120}{2} = 60$$

∴ 60

4

How many handshakes occur in a room of 15 people if everyone shakes hands with everyone else exactly once?

Correct!

Incorrect. Correct: 105

Solution

Every handshake involves exactly 2 people. Since 'A shakes B's hand' is identical to 'B shakes A's hand', order doesn't matter → use Combination.

Select 2 people from 15:
$$C(15,2) = \frac{15 \times 14}{2 \times 1} = 105$$

∴ 105

5

The value of C(10, 4) + C(10, 5) equals:

Correct!

Incorrect. Correct: C(11, 5)

Solution

Recall Pascal's Rule:
$$C(n,\,r-1) + C(n,\,r) = C(n+1,\,r)$$

Match to given expression: $C(10,4) + C(10,5)$.
Here $n = 10$ and $r = 5$ (so $r-1 = 4$).

Apply the rule:
$$C(10,4) + C(10,5) = C(10+1,\; 5) = C(11,5)$$

∴ C(11, 5)

6

How many 3-digit numbers can be formed using digits 1, 2, 3 if repetition is allowed?

Correct!

Incorrect. Correct: 27

Solution

Repetition is allowed, so each position is independent and has all 3 choices available.

Hundreds place: 3 choices.
Tens place: 3 choices (repetition allowed, full 3 still available).
Units place: 3 choices.

$$3 \times 3 \times 3 = 3^3 = 27$$

∴ 27

7

Out of 8 consonants and 4 vowels, how many ways can we select 3 consonants and 2 vowels?

Correct!

Incorrect. Correct: 336

Solution

Two independent selection tasks:

Select 3 consonants from 8:
$$C(8,3) = \frac{8 \times 7 \times 6}{3!} = 56$$

Select 2 vowels from 4:
$$C(4,2) = \frac{4 \times 3}{2!} = 6$$

Both selections happen together (AND) → multiply:
$$56 \times 6 = 336$$

∴ 336

8

You want to seat 5 people at a round table where A and B must sit together. How many ways?

Correct!

Incorrect. Correct: 12

Solution

Bundle Trick: A and B must be adjacent → treat them as one unit (AB).

Now 4 units to arrange in a circle (AB-bundle + C + D + E):
$$(4-1)! = 3! = 6 \text{ ways}$$

Inside the bundle, A and B can be (AB) or (BA):
$$2! = 2 \text{ ways}$$

Multiply:
$$6 \times 2 = 12$$

∴ 12

9

There are 4 math books and 3 science books. If you choose exactly 1 book to read, how many choices?

Correct!

Incorrect. Correct: 7

Solution

You pick exactly 1 book. You either pick from math OR from science — these are mutually exclusive (you can't do both at once).

Apply the Addition Principle:
$$4 + 3 = 7 \text{ choices}$$

∴ 7

10

The value of 0! is:

Correct!

Incorrect. Correct: 1

Solution

$0!$ is defined as 1 by mathematical convention.

One way to see why: using the factorial recurrence $n! = n \times (n-1)!$, we can write $1! = 1 \times 0!$, so $0! = \frac{1!}{1} = 1$.

Another reason: $P(n,n) = \frac{n!}{(n-n)!} = \frac{n!}{0!}$ should equal $n!$ (arranging all n items). This only works if $0! = 1$.

∴ 1

Short Questions (2 Marks Each)

1

Find the value of n if P(n, 3) = 6 × P(n, 2).

[2 marks]

Solution

Expand both sides using $P(n,r) = \frac{n!}{(n-r)!}$:

$P(n,3) = n(n-1)(n-2)$
$P(n,2) = n(n-1)$

Substitute into the equation:
$$n(n-1)(n-2) = 6 \times n(n-1)$$

Since $n \geq 3$ (otherwise $P(n,3)$ is undefined), divide both sides by the common factor $n(n-1)$:
$$n - 2 = 6$$
$$n = 8$$

∴ n = 8

2

In how many ways can 5 boys and 4 girls be arranged in a straight line so that no two girls are adjacent?

[2 marks]

Solution

Use the Gaps Trick: arrange the boys first, then fill the gaps with girls.

Step 1 — Arrange 5 boys in a line:
$$5! = 120 \text{ ways}$$

Step 2 — Count the gaps created:
_ B _ B _ B _ B _ B _ → 6 gaps

Step 3 — Place 4 girls in 6 gaps (one girl per gap, so no two are adjacent):
$$P(6,4) = 6 \times 5 \times 4 \times 3 = 360 \text{ ways}$$

Step 4 — Multiply (both steps happen together):
$$120 \times 360 = 43{,}200$$

∴ 43,200 arrangements

3

In how many ways can 8 people be seated around a circular table?

[1 marks]

Solution

Circular arrangement: fix one person's position as a reference point (removes rotational symmetry). The remaining 7 people fill the remaining 7 seats.

$$\text{Circular arrangements} = (n-1)! = (8-1)! = 7!$$
$$7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040$$

∴ 5,040 ways

4

A box contains 5 red balls and 6 black balls. In how many ways can 3 balls of the same colour be drawn?

[2 marks]

Solution

'Same colour' means either all 3 red OR all 3 black — two mutually exclusive events.

All 3 red (from 5):
$$C(5,3) = 10$$

All 3 black (from 6):
$$C(6,3) = 20$$

Since these are OR-events, apply the Addition Principle:
$$10 + 20 = 30$$

∴ 30 ways

5

How many 4-digit numbers can be formed using digits 0, 2, 4, 6, 8 without repetition?

[2 marks]

Solution

The thousands digit cannot be 0 (otherwise it's not a 4-digit number).

Thousands place: Choose from {2, 4, 6, 8} → 4 choices.

Remaining 3 places: Now 4 digits remain (including 0), fill 3 slots without repetition:
$$P(4,3) = 4 \times 3 \times 2 = 24 \text{ ways}$$

Multiply:
$$4 \times 24 = 96$$

∴ 96 numbers

6

Find the number of diagonals in a decagon (10-sided polygon).

[2 marks]

Solution

Any 2 vertices of a decagon form a straight line. Total lines from 10 vertices:
$$C(10,2) = \frac{10 \times 9}{2} = 45 \text{ lines}$$

But 10 of these lines are the sides of the polygon (not diagonals). Subtract them:
$$45 - 10 = 35 \text{ diagonals}$$

General formula for any n-gon: $\text{Diagonals} = C(n,2) - n = \frac{n(n-3)}{2}$

∴ 35 diagonals

7

If C(15, r) = C(15, 2r − 3), find the value of r.

[2 marks]

Solution

If $C(n,a) = C(n,b)$, then either $a = b$ OR $a + b = n$.

Case 1 — Equal indices:
$r = 2r - 3 \Rightarrow r = 3$

Case 2 — Complementary (sum = 15):
$r + (2r-3) = 15$
$3r - 3 = 15$
$r = 6$

Both are valid (check: $C(15,3) = C(15,12)$ ✓ and $C(15,6) = C(15,9)$ ✓).

∴ r = 3 or r = 6

8

Prove that C(n, r) = P(n, r) / r!

[2 marks]

Solution

Consider selecting r items from n and arranging them.

$C(n,r)$ = number of ways to choose a group of r (order irrelevant).

For any one chosen group, the r items inside it can be arranged in $r!$ ways.

So: choosing a group AND arranging it = $C(n,r) \times r!$ total ordered arrangements.

But arranging r items chosen from n in order is exactly $P(n,r)$:
$$C(n,r) \times r! = P(n,r)$$

Divide both sides by $r!$:
$$C(n,r) = \frac{P(n,r)}{r!} \quad \blacksquare$$

∴ Proved: C(n,r) = P(n,r) / r!

9

In how many ways can you drop 4 distinct letters into 6 different mailboxes?

[1 marks]

Solution

Each letter can go into any of the 6 mailboxes independently — multiple letters can share a mailbox (repetition allowed).

This is a repetition problem with $n = 6$ mailboxes and $r = 4$ letters:
$$6^4 = 6 \times 6 \times 6 \times 6 = 1296$$

∴ 1,296 ways

10

How many words can be formed using all letters of 'ENGINEERING'?

[2 marks]

Solution

Count letters and their frequencies in ENGINEERING:
Total = 11 letters
E: 3 times N: 3 times G: 2 times I: 2 times R: 1 time

Identical objects formula:
$$\frac{11!}{3! \times 3! \times 2! \times 2!} = \frac{39{,}916{,}800}{6 \times 6 \times 2 \times 2} = \frac{39{,}916{,}800}{144}$$

∴ 277,200 words

Long Questions (3 Marks Each)

1

In how many ways can the letters of 'MATHEMATICS' be arranged so that all vowels always come together?

[3 marks]

Solution

Identify vowels in MATHEMATICS: A, E, A, I (4 vowels; A repeats 2 times).
Consonants: M, T, H, M, T, C, S (7 consonants; M repeats 2×, T repeats 2×).

Step 1 — Bundle the 4 vowels into one unit [AEAI].
Now arrange 8 units: [AEAI] + M + T + H + M + T + C + S.
With M repeating 2× and T repeating 2×:
$$\frac{8!}{2! \times 2!} = \frac{40320}{4} = 10{,}080 \text{ ways}$$

Step 2 — Arrange letters inside the vowel bundle [AEAI]:
4 letters, A repeating 2×:
$$\frac{4!}{2!} = \frac{24}{2} = 12 \text{ ways}$$

Step 3 — Multiply:
$$10{,}080 \times 12 = 120{,}960$$

∴ 120,960 ways

2

From 6 teachers and 8 students, a committee of 5 must include at least 2 teachers. How many ways?

[3 marks]

Solution

'At least 2 teachers' → split into 4 cases and add (Addition Principle):

Case 1: 2 teachers + 3 students
$$C(6,2) \times C(8,3) = 15 \times 56 = 840$$

Case 2: 3 teachers + 2 students
$$C(6,3) \times C(8,2) = 20 \times 28 = 560$$

Case 3: 4 teachers + 1 student
$$C(6,4) \times C(8,1) = 15 \times 8 = 120$$

Case 4: 5 teachers + 0 students
$$C(6,5) \times C(8,0) = 6 \times 1 = 6$$

Total:
$$840 + 560 + 120 + 6 = 1526$$

∴ 1,526 ways

3

Find the number of ways 5 gentlemen and 5 ladies can be seated at a round table so that they alternate.

[3 marks]

Solution

For alternating seating in a circle, no two of the same gender can be adjacent.

Step 1 — Fix the gentlemen in a circle:
$(5-1)! = 4! = 24$ ways

Step 2 — Count gaps between the seated gentlemen:
5 gentlemen in a circle create exactly 5 gaps between them (one between each adjacent pair). These gaps are now distinct, fixed positions.

Step 3 — Seat 5 ladies in 5 fixed gaps:
$5! = 120$ ways (linear, since the gaps are already positioned)

Multiply:
$$24 \times 120 = 2880$$

∴ 2,880 ways

4

In how many ways can the letters of 'SOCIOLOGICAL' be arranged so that all 3 O's never come together?

[3 marks]

Solution

Strategy: Total arrangements − Arrangements where all 3 O's ARE together.

Total letters in SOCIOLOGICAL = 12
O: 3 times C: 2 times I: 2 times L: 2 times

Total unrestricted arrangements:
$$\frac{12!}{3! \times 2! \times 2! \times 2!} = \frac{479{,}001{,}600}{48} = 9{,}979{,}200$$

Unfavourable (all 3 O's bundled as [OOO]):
Now 10 units to arrange: [OOO] + 9 other letters, with C, I, L each still repeating 2×:
$$\frac{10!}{2! \times 2! \times 2!} = \frac{3{,}628{,}800}{8} = 453{,}600$$

Favourable = Total − Unfavourable:
$$9{,}979{,}200 - 453{,}600 = 9{,}525{,}600$$

∴ 9,525,600 ways

5

An exam has 12 questions: Section A has 7, Section B has 5. A student must attempt 8 total with at least 3 from each section. How many ways?

[3 marks]

Solution

Must choose 8 total with at least 3 from A AND at least 3 from B. Enumerate valid combinations:

Option 1: 3 from A + 5 from B
$$C(7,3) \times C(5,5) = 35 \times 1 = 35$$

Option 2: 4 from A + 4 from B
$$C(7,4) \times C(5,4) = 35 \times 5 = 175$$

Option 3: 5 from A + 3 from B
$$C(7,5) \times C(5,3) = 21 \times 10 = 210$$

(Note: 6 from A + 2 from B violates 'at least 3 from B'. Similarly 2+6 violates 'at least 3 from A'.)

Total:
$$35 + 175 + 210 = 420$$

∴ 420 ways

6

If P(n, r) = 2520 and C(n, r) = 21, find n and r.

[3 marks]

Solution

Use the fundamental relationship $C(n,r) = \frac{P(n,r)}{r!}$:

Step 1 — Find r:
$$21 = \frac{2520}{r!} \Rightarrow r! = \frac{2520}{21} = 120$$
Since $5! = 120$, we get $r = 5$.

Step 2 — Find n using P(n, 5) = 2520:
$$n(n-1)(n-2)(n-3)(n-4) = 2520$$

Try $n = 7$: $7 \times 6 \times 5 \times 4 \times 3 = 2520$ ✓

∴ n = 7, r = 5

7

How many 5-digit odd numbers can be formed from 1, 2, 3, 4, 5, 6, 7 without repetition?

[3 marks]

Solution

For the number to be odd, the units digit must be odd.
Odd digits from {1,2,3,4,5,6,7}: 1, 3, 5, 7 → 4 choices for units digit.

After fixing the units digit, 6 digits remain for the other 4 positions (no repetition):
$$P(6,4) = 6 \times 5 \times 4 \times 3 = 360 \text{ ways}$$

Multiply:
$$4 \times 360 = 1440$$

∴ 1,440 numbers

8

A committee of 6 is formed from 7 men and 5 women with exactly 4 men. How many ways?

[3 marks]

Solution

Exactly 4 men in a 6-member committee means exactly $6 - 4 = 2$ women.

Select 4 men from 7:
$$C(7,4) = 35$$

Select 2 women from 5:
$$C(5,2) = 10$$

Both selections happen together (AND) → multiply:
$$35 \times 10 = 350$$

∴ 350 ways

9

Prove Pascal's Rule algebraically: C(n, r−1) + C(n, r) = C(n+1, r).

[3 marks]

Solution

Expand the left-hand side using the factorial definition:
$$\frac{n!}{(r-1)!(n-r+1)!} + \frac{n!}{r!(n-r)!}$$

Find the common denominator $= r!(n-r+1)!$. Multiply the 1st fraction by $\frac{r}{r}$ and the 2nd by $\frac{(n-r+1)}{(n-r+1)}$:
$$\frac{n! \cdot r}{r!(n-r+1)!} + \frac{n! \cdot (n-r+1)}{r!(n-r+1)!}$$

Combine over the common denominator:
$$\frac{n![r + (n-r+1)]}{r!(n-r+1)!} = \frac{n!(n+1)}{r!(n-r+1)!}$$

Since $n!(n+1) = (n+1)!$ and $(n-r+1) = (n+1)-r$:
$$= \frac{(n+1)!}{r!((n+1)-r)!} = C(n+1,r) \quad \blacksquare$$

∴ Algebraically proved

10

A cricket squad of 15: captain and wicket-keeper must be included; 2 injured players must be excluded. Select 11 players. How many ways?

[3 marks]

Solution

Apply constraints step by step:

Step 1 — Remove 2 injured players:
Available pool: $15 - 2 = 13$ players.

Step 2 — Fix captain and wicket-keeper as mandatory:
They occupy 2 spots, leaving $11 - 2 = 9$ spots to fill.
Remaining pool: $13 - 2 = 11$ players to choose from.

Step 3 — Select 9 from 11:
$$C(11,9) = C(11,2) = \frac{11 \times 10}{2} = 55$$

∴ 55 ways

Past NEB Board Questions

1

Find the number of ways 5 boys and 4 girls can be arranged in a row so that no two girls sit together.

NEB 2079 [3 marks]

Solution

Step 1 — Arrange 5 boys in a straight line:
$$5! = 120 \text{ ways}$$

Step 2 — Identify gaps:
_ B _ B _ B _ B _ B _ → 6 gaps available

Step 3 — Place 4 girls into 6 gaps (one per gap ensures no two girls are adjacent):
$$P(6,4) = 6 \times 5 \times 4 \times 3 = 360 \text{ ways}$$

Total:
$$120 \times 360 = 43{,}200$$

∴ 43,200 ways

2

Prove that C(n, r−1) + C(n, r) = C(n+1, r). [Combinatorial Proof]

NEB 2078 [3 marks]

Solution

Combinatorial (Committee) Proof:

Consider choosing a committee of $r$ people from $(n+1)$ people. This can be done in $C(n+1,r)$ ways — our target.

Label one specific person as Person X. Every committee either includes X or excludes X. These two cases are mutually exclusive and exhaustive.

Case 1 — X is included:
Choose the remaining $(r-1)$ members from the other $n$ people:
$C(n, r-1)$ ways.

Case 2 — X is excluded:
Choose all $r$ members from the other $n$ people:
$C(n, r)$ ways.

Adding both cases (no overlap):
$$C(n, r-1) + C(n, r) = C(n+1, r) \qquad \blacksquare$$

∴ Combinatorial proof complete

3

In how many ways can the letters of 'MATHEMATICS' be arranged so that all vowels come together?

NEB 2077 [3 marks]

Solution

Vowels: A, E, A, I (4 vowels, A repeats 2×). Consonants: M, T, H, M, T, C, S (7; M×2, T×2).

Bundle vowels as [AEAI]: Now 8 units.

Arrange 8 units (with M×2, T×2 repeating):
$$\frac{8!}{2! \times 2!} = 10{,}080$$

Internal arrangements of vowel bundle (4 letters, A×2):
$$\frac{4!}{2!} = 12$$

Total:
$$10{,}080 \times 12 = 120{,}960$$

∴ 120,960 ways

Chapter Test — Full Mixed Paper

Attempt this test in 45 minutes without looking at notes. Then check your solutions.

1 Section A — MCQs (1 Mark Each)

Mixed Level — MCQ

1

The value of 0! is:

Correct!

Incorrect. Correct: 1

Solution

By convention and to keep formulas like $P(n,n) = n!/0!$ consistent:
$$0! = 1$$

∴ 1

2

In how many ways can 4 people sit around a circular table?

Correct!

Incorrect. Correct: 6

Solution

Circular arrangement of n people = $(n-1)!$
$$( 4 - 1 )! = 3! = 6$$

∴ 6

3

C(10, 3) equals:

Correct!

Incorrect. Correct: 120

Solution

$$C(10,3) = \frac{10!}{3! \times 7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = \frac{720}{6} = 120$$

∴ 120

2 Section B — Short Questions (2 Marks Each)

1

How many diagonals does a decagon (10 sides) have?

[2 marks]

Solution

Total lines from 10 vertices = $C(10,2) = 45$.
Subtract the 10 sides of the polygon:
$$45 - 10 = 35 \text{ diagonals}$$

∴ 35

2

Find n if C(n, 2) = 21.

[2 marks]

Solution

$$C(n,2) = \frac{n(n-1)}{2} = 21$$
$$n(n-1) = 42$$
Testing: $7 \times 6 = 42$ ✓
$$n = 7$$

∴ n = 7

3 Section C — Long Questions (3 Marks Each)

1

Find the number of ways 5 gentlemen and 5 ladies can be seated at a round table so that they alternate.

[3 marks]

Solution

Step 1 — Fix 5 gentlemen in a circle:
$(5-1)! = 24$ ways.

Step 2 — 5 gaps are created between them. Place 5 ladies in 5 distinct gaps:
$5! = 120$ ways.

Total: $24 \times 120 = 2880$

∴ 2,880

2

In how many ways can 6 men and 4 women be arranged in a row so that no two women are adjacent?

[3 marks]

Solution

Step 1 — Arrange 6 men in a line:
$6! = 720$ ways.

Step 2 — Count gaps:
_ M _ M _ M _ M _ M _ M _ → 7 gaps.

Step 3 — Place 4 women in 7 gaps:
$P(7,4) = 7 \times 6 \times 5 \times 4 = 840$ ways.

Total:
$$720 \times 840 = 604{,}800$$

∴ 604,800

One-Page Revision Cheat Sheet

1

Swap test first: Swapping chosen items changes outcome? → P. Doesn't change? → C.

2

Repetition allowed? → Use $n^r$ (codes, PINs, mailboxes).

3

Linear arrangement: $P(n,r) = n!/(n-r)!$ — order matters, no repeats.

4

Identical letters/objects: $n! \div (p! \times q! \times \cdots)$ — divide by each repeat group.

5

Round table (fixed): $(n-1)!$ — fix one person, arrange the rest.

6

Necklace / Bracelet (flippable): $\frac{1}{2}(n-1)!$ — divide by 2 extra.

7

Selection (group): $C(n,r) = n! \div [r!(n-r)!]$ — order doesn't matter.

8

Complementary shortcut: $C(n,r) = C(n,n-r)$ — use when r is large.

9

Pascal's Rule: $C(n,r-1) + C(n,r) = C(n+1,r)$.

10

Together trick: Bundle into 1 block → arrange blocks → × internal arrangements.

11

Never together: Total − Together.

12

No two adjacent / alternate: Seat larger group first → count gaps → fill gaps with P.

13

At least / at most: Split into separate cases, add with Addition Principle.

14

AND → multiply. OR (exclusive) → add.

15

Diagonals of n-gon: $C(n,2) - n = n(n-3)/2$. Handshakes in n people: $C(n,2)$.