📘 Algebra · Chapter MN05🎯 NDA Level : High Priority
Permutations and Combinations (P&C) is one of the most reliable scoring chapters in NDA Mathematics. Questions are straightforward once the core difference between "order matters" (permutations) and "order does not matter" (combinations) is clear. The 2023–2025 NDA papers consistently tested nPr, nCr, selections with restrictions, and division into groups.
📌 What to expect in NDA (based on 2022–2024 papers): (1) Direct evaluation of ⁿPᵣ and ⁿCᵣ for given n and r;
(2) Solving equations like ⁿCᵣ = ⁿCₛ → r+s = n;
(3) Arrangements with restrictions — certain people always/never together;
(4) Circular permutations;
(5) Words from letters — with or without repetition;
(6) Selecting a committee with conditions (at least one woman, etc.);
(7) Number of diagonals in a polygon;
(8) Division into groups (equal and unequal).
Topics at a Glance
① Counting Principles
Addition rule, multiplication rule
② Factorial & ⁿPᵣ
n!, ⁿPᵣ = n!/(n−r)!, restrictions
③ Circular Permutations
(n−1)!, necklace, beads
④ Permutations with Repetition
Identical items, repeated letters
⑤ Combinations ⁿCᵣ
n!/r!(n−r)!, properties, groups
⑥ Applications
Committee, polygon, selection problems
1. Fundamental Principles of Counting
1.1
Addition Principle & Multiplication Principle
These two rules underlie every counting problem — master them first
➕ Addition Principle (OR rule)
If a task can be done in m ways OR another mutually exclusive task in n ways, then either task can be done in m + n ways
Key word: OR → ADD
Conditions: the two events must be mutually exclusive (cannot happen at the same time)
Example: Travel by bus (3 routes) OR train (5 routes) → 3+5 = 8 total ways
✖ Multiplication Principle (AND rule)
If task 1 can be done in m ways AND task 2 (independent) in n ways, then both together can be done in m × n ways
Worked Example — Multiplication Principle (Boxes Method)
How many 3-digit numbers can be formed using digits 1–9 with no repetition?
Think of 3 boxes, one for each digit position:
9
Hundreds
×
8
Tens
×
7
Units
=
504
9 choices for hundreds (1–9) × 8 remaining for tens × 7 remaining for units = 504.
📌 The Boxes Method (Slot Method):
Draw one box for each position or choice to be made. Fill in how many options are available for each box, then multiply all boxes together. This works for virtually every NDA permutation/combination word problem and makes the logic visual and error-free.
2. Factorial Notation & Permutations (ⁿPᵣ)
2.1
Factorial & Linear Permutations
Order matters in permutations — arrangement is the key word
A permutation is an arrangement of objects where order matters. Selecting r objects from n and arranging them gives ⁿPᵣ arrangements.
⚡ Factorial & Permutation Formulae
Factorial: n! = n × (n−1) × (n−2) × … × 2 × 1
0! = 1 (by definition)
1! = 1
Permutation: ⁿPᵣ = n! / (n−r)! where 0 ≤ r ≤ n
Special cases:
ⁿP₀ = 1 (arranging 0 items — one way: do nothing)
ⁿP₁ = n (choose 1 from n, arrange it)
ⁿPₙ = n! (arrange all n items)
ⁿPᵣ = n · (n−1) · (n−2) · … · (n−r+1) ← r factors
Restrictions are the most tested permutation type in NDA
Restriction Type
Method
Formula / Approach
Certain items ALWAYS together
Treat them as one block, then arrange block internally
Arrangements = (n−k+1)! × k! (k items together)
Certain items NEVER together
Total arrangements − (arrangements where they ARE together)
n! − (n−k+1)! × k!
Fixed positions (e.g., ends fixed)
Fix those positions first, then fill the rest
Fill fixed: one way. Remaining (n−fixed)! ways.
Odd/even positions separate
Assign one group to odd positions, other to even
Odd positions: ⌈n/2⌉!, Even positions: ⌊n/2⌋!
Relative order fixed
Arrange all, divide by number of identical orderings
n! / k! (if k identical items)
Worked Example — Two People Always Together
In how many ways can 6 people be arranged in a row so that A and B are always together?
Treat A and B as one block → 5 units to arrange: 5! ways.
Within the block, A and B can swap: 2! = 2 ways.
Total = 5! × 2! = 120 × 2 = 240.
Worked Example — Two People Never Together
In how many ways can 6 people be arranged in a row so that A and B are never together?
Total arrangements = 6! = 720.
Arrangements where A and B ARE together = 240 (from above).
Never together = 720 − 240 = 480.
2.3
Circular Permutations
One position is fixed to remove rotational equivalence
In a circular arrangement, rotations of the same arrangement are considered identical. So one position is "fixed" and the remaining n−1 are arranged.
⚡ Circular Permutation Formulae
Circular arrangements of n distinct objects = (n − 1)!
Necklace / Bracelet (can be flipped — clock = anti-clock):
= (n − 1)! / 2
n people around a circular table:
= (n − 1)!
n beads on a necklace:
= (n − 1)! / 2
Why (n−1)!? In a line, n! arrangements. In a circle, each unique arrangement appears n times (once for each rotation). So divide by n: n!/n = (n−1)!.
Fig 1: Linear vs Circular arrangement — A is fixed in the circle to eliminate rotational equivalence; only B, C, D are freely arranged.
2.4
Permutations with Identical (Repeated) Objects
Divide by factorial of each group of identical items
⚡ Formula — Arrangements with Identical Items
If n items have p identical of one kind, q of another, r of a third, …:
Number of arrangements = n! / (p! × q! × r! × …)
Examples:
Letters of MISSISSIPPI (11 letters: M=1, I=4, S=4, P=2):
= 11! / (1! × 4! × 4! × 2!) = 34650
Letters of MATHEMATICS (11 letters: M=2, A=2, T=2, others=1 each):
= 11! / (2! × 2! × 2!) = 4989600
This is one of the most tested sub-topics in NDA — "how many distinct words can be formed from the letters of …?" Always count identical letters carefully before substituting.
Worked Example — Words from ARRANGE
How many distinct arrangements of the letters of ARRANGE are possible?
Q5. How many 3-digit even numbers can be formed from 1, 2, 3, 4, 5 without repetition?
(a) 24 (b) 36 (c) 48 (d) 60
Answer: (a) 24
Units digit must be 2 or 4 → 2 choices.
Hundreds: 4 remaining choices. Tens: 3 remaining.
Total = 4 × 3 × 2 = 24.
🔥 TRICKY QUESTIONS
Permutations — Classic NDA Traps
🧩 T1. In how many ways can the letters of INDIA be arranged so that the two I's are never together?
INDIA: I=2, N=1, D=1, A=1. Total letters = 5.
Total distinct arrangements = 5!/2! = 60.
Arrangements with both I's together: treat II as one unit → 4 letters → 4! = 24.
Never together = 60 − 24 = 36. Trap: Using 5! instead of 5!/2! for total — forgetting the two identical I's makes some arrangements non-distinct.
🧩 T2. How many numbers greater than 1000 can be formed using digits 0, 1, 2, 3 (each used once)?
Numbers > 1000 must be 4-digit → arrange all 4 digits: 4! = 24.
But numbers starting with 0 are invalid (3-digit): fix 0 at thousands place → 3! = 6 such cases.
Valid numbers = 24 − 6 = 18. Trap: Counting 0 as a valid leading digit. Always check for the leading-zero restriction in digit problems.
🧩 T3. How many arrangements of MATHEMATICS begin and end with a vowel?
MATHEMATICS: M=2, A=2, T=2, H=1, E=1, I=1, C=1, S=1. Total 11 letters, 4 vowels (A, A, E, I).
Choose first position (vowel): options from {A, A, E, I}. Choose last position (vowel, different slot).
Case 1: Both A's at ends (end–end = A–A): middle 9 letters (M=2,T=2,others) = 9!/(2!×2!) = 90720.
Case 2: One A at one end, E or I at other (2 choices for which end gets A, 2 choices for E/I): 2×2 × 9!/(2!×2!) = 4 × 90720 = 362880. But correct method — treat systematically using cases. This is a higher-difficulty NDA problem. The approach: fix the vowels at ends, count internal arrangements, multiply by valid end-pair choices.
3. Combinations (ⁿCᵣ)
3.1
Definition, Formula & Properties of ⁿCᵣ
Selection only — order does not matter here
A combination is a selection of objects where order does not matter. Choosing r objects from n without caring about arrangement gives ⁿCᵣ ways.
⚡ Combination Formula & Properties
ⁿCᵣ = n! / [r! × (n−r)!] also written C(n,r) or (n choose r)
Relation to permutation: ⁿCᵣ = ⁿPᵣ / r!
Key Properties:
ⁿC₀ = 1 ⁿCₙ = 1 ⁿC₁ = n
ⁿCᵣ = ⁿCₙ₋ᵣ (symmetry — very useful for large r)
If ⁿCᵣ = ⁿCₛ then r = s or r + s = n
ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ (Pascal's identity)
Σ ⁿCᵣ (r=0 to n) = 2ⁿ (total subsets of n items)
Distinguish between named groups (labelled) and unnamed groups (unlabelled)
⚡ Division into Groups — All Cases
Case 1 — Divide n items into UNEQUAL groups of p, q, r (p+q+r=n):
Ways = n! / (p! × q! × r!) [groups are distinguishable by size]
Case 2 — Divide n items into EQUAL groups of k each, groups NOT labelled:
Ways = n! / [(k!)^m × m!] where m = n/k groups
(divide by m! because the groups themselves are interchangeable)
Case 3 — Divide n items into EQUAL groups of k each, groups ARE labelled:
Ways = n! / (k!)^m (no division by m! since groups are distinct)
Example — 6 persons into 3 labelled pairs:
= 6! / (2!)³ = 720 / 8 = 90
Example — 6 persons into 3 unlabelled pairs:
= 6! / [(2!)³ × 3!] = 720 / 48 = 15
The key question: are the groups themselves distinguishable? If groups have names/labels/rooms → labelled. If groups are interchangeable (just "groups") → unlabelled, so divide by m! at the end.
⚠ Most Tested Trap — Equal Groups:
The most common error in division-into-groups problems is forgetting to divide by m! when the groups are unlabelled and equal in size. If the problem says "divide into groups of 3 each" (no names), divide by 3! = 6 extra. If it says "put into three rooms of 3 each", the rooms are labelled — do not divide by 3!.
3.4
Standard Applications of ⁿCᵣ
Formulae for diagonals, handshakes, triangles — memorise these results
Q9. From 12 students, a committee of 5 is to be formed. In how many ways can it be done if a particular student is always included?
(a) 165 (b) 330 (c) 462 (d) 792
Answer: (b) 330
The particular student is fixed. Choose remaining 4 from 11 students.
¹¹C₄ = 11×10×9×8/(4×3×2×1) = 7920/24 = 330.
Q10. How many triangles can be formed from 10 points on a circle?
(a) 100 (b) 110 (c) 120 (d) 90
Answer: (c) 120
Points on a circle are all non-collinear (any 3 form a triangle).
¹⁰C₃ = 10×9×8/(3×2×1) = 120.
🔥 TRICKY QUESTIONS
Combinations — Common NDA Traps
🧩 T4. In how many ways can a cricket team of 11 be selected from 15 players if a particular player must NOT be selected and another player must ALWAYS be selected?
One player fixed in (always selected) → select 10 more from remaining 15−2 = 13 players (excluding both the forced-in and forced-out players).
Ways = ¹³C₁₀ = ¹³C₃ = 13×12×11/(3×2×1) = 286. Trap: Students subtract restrictions one at a time instead of applying both simultaneously. Identify all fixed/excluded players first, then choose from what remains.
🧩 T5. How many 4-letter words (with or without meaning) can be formed using the letters of LOGARITHMS, if each word contains the letter O?
LOGARITHMS has 10 distinct letters (no repeats). O must be in the word.
Fix O in the word → choose 3 more letters from remaining 9: ⁹C₃ ways.
Now arrange all 4 letters (including O): 4! ways.
Total = ⁹C₃ × 4! = 84 × 24 = 2016. Trap: Selecting the letters first and forgetting to multiply by 4! for arrangements (this is a word arrangement, not just a selection).
🧩 T6. There are 12 points in a plane, 5 of which are collinear. How many distinct straight lines can be drawn?
Without restriction: ¹²C₂ = 66 lines.
The 5 collinear points give ⁵C₂ = 10 "lines" but they all lie on 1 actual line.
Distinct lines = 66 − 10 + 1 = 57. Key formula: Lines = ⁿC₂ − ᵐC₂ + 1 (subtract the m(m−1)/2 pairs among collinear points and add back the 1 actual line they form).
🧩 T7. In how many ways can 9 books be divided equally among 3 students?
3 books each, 3 named students (labelled groups).
Student A gets 3 books: ⁹C₃ = 84 ways.
Student B gets 3 from remaining 6: ⁶C₃ = 20 ways.
Student C gets remaining 3: ¹C₃ → ³C₃ = 1 way.
Total = 84 × 20 × 1 = 1680.
If students were NOT named (groups unlabelled): 1680/3! = 280. Always ask: are the groups labelled? Here yes (named students), so no extra division by 3!.
📋 Master Formula Sheet — MN05 Permutations & Combinations
All critical formulae for rapid pre-exam revision.
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