Permutations and Combinations

P&C gives NDA 2–4 questions and also feeds probability and the binomial theorem. The whole topic rests on one distinction — order matters (permutation) or not (combination) — plus a handful of formulas.

Core ideas

Fundamental counting principle: if a job has m ways and another n ways, together they can be done in m×n ways (AND) or m+n ways (OR, mutually exclusive).

Permutations (order matters): ⁿPᵣ = n!/(n−r)!. Arrangements of n distinct objects = n!.

Combinations (order irrelevant): ⁿCᵣ = n!/[r!(n−r)!]. Note ⁿPᵣ = ⁿCᵣ × r!.

Key identities: ⁿCᵣ = ⁿCₙ₋ᵣ; ⁿC₀ = ⁿCₙ = 1; ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ (Pascal's rule).

Repetitions: arrangements of n objects with repeats p, q, r alike = n!/(p! q! r!). Circular permutations of n distinct objects = (n−1)!, and (n−1)!/2 if clockwise and anticlockwise are the same (like a necklace).

Exam Tricks & Tips

• 🎯 Decide FIRST whether order matters — 'arrange/word/rank' ⇒ permutation; 'select/team/committee' ⇒ combination.
• 🎯 'At least one' problems: use total − none. Number of ways to select at least one from n distinct = 2ⁿ − 1.
• 🎯 For 'always together', glue the group as one unit, arrange, then multiply by internal arrangements.
• 🎯 For 'never together', subtract the together-count from the total.
• 🎯 ⁿCᵣ is maximum at r = n/2 (middle) — useful when a question asks which term is largest.
• ❌ Common mistake: applying ⁿPᵣ to a committee selection. A committee has no order, so it is ⁿCᵣ, not ⁿPᵣ.

Expected exam pattern

Expect a word-arrangement (with repeated letters), a committee/handshake selection, a 'digits forming numbers' problem, or a circular seating. Handshakes among n people = ⁿC₂.

Quick recap

Separate order (P) from no-order (C). Memorise ⁿPᵣ, ⁿCᵣ, the symmetry ⁿCᵣ=ⁿCₙ₋ᵣ, Pascal's rule, repetition formula n!/(p!q!r!), and circular (n−1)!. Use total-minus-complement for 'at least' problems.

Cover, recall, check. 11 cards on counting.

Q1. ⁿPᵣ formula?
A1. n!/(n−r)!.

Q2. ⁿCᵣ formula?
A2. n!/[r!(n−r)!].

Q3. Relation between ⁿPᵣ and ⁿCᵣ?
A3. ⁿPᵣ = ⁿCᵣ × r!.

Q4. Symmetry property of combinations?
A4. ⁿCᵣ = ⁿCₙ₋ᵣ.

Q5. Pascal's rule?
A5. ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ.

Q6. Arrangements of n items with p, q, r alike?
A6. n!/(p! q! r!).

Q7. Circular permutations of n distinct objects?
A7. (n−1)!; for a necklace/garland it is (n−1)!/2.

Q8. Number of ways to select at least one from n distinct objects?
A8. 2ⁿ − 1.

Q9. Number of handshakes among n people?
A9. ⁿC₂ = n(n−1)/2.

Q10. 'Always together' method?
A10. Treat the group as one unit, arrange, then multiply by internal arrangements.

Q11. Sum of all ⁿCᵣ from r=0 to n?
A11. 2ⁿ.

Counting without listing — that is the whole art of permutations and combinations, and it is the engine room of NDA's probability questions.

Core idea / what this tests: Deciding whether order matters (permutation) or not (combination), and handling restrictions like repetition, identical objects and circular arrangements.

Deep explanation

Beginner — The fundamental principle

If a job has m outcomes and a second independent job has n outcomes, together they have m × n outcomes (multiplication principle). "Or" situations add; "and" situations multiply.
Factorial: n! = n·(n−1)···2·1, with 0! = 1.

Intermediate — Permutations vs combinations

• Permutation (order matters): ⁿPᵣ = n! / (n−r)!.
• Combination (order does not): ⁿCᵣ = n! / [r!(n−r)!].
  Link: ⁿPᵣ = ⁿCᵣ · r!. Also ⁿCᵣ = ⁿC₍ₙ₋ᵣ₎ and ⁿC₀ = ⁿCₙ = 1.

Advanced — Special cases

• Identical objects: arrangements of n items with repeats p, q, … = n! / (p!·q!·…).
• Circular permutations: (n−1)! arrangements; ÷2 more if reflections (necklace) coincide.
• With repetition allowed: number of r-length strings from n symbols = nʳ.
• At least / at most: use complementary counting (total − unwanted).

Worked example

In how many ways can the letters of the word BALLOON be arranged?
BALLOON has 7 letters with L repeated twice and O repeated twice. Arrangements = 7! / (2!·2!) = 5040 / 4 = 1260.

NDA relevance

Direct word-arrangement and committee-selection MCQs appear every year, and P&C results feed straight into probability numerators/denominators. Circular-arrangement and "digits forming numbers" problems are recurring twists.

Exam tricks & shortcuts

• Decide first: does swapping two chosen items make a new outcome? Yes → permutation; No → combination.
• "At least one" is almost always faster as 1 − P(none) or Total − none.
• Keep required objects together by treating them as a single block, then multiply by internal arrangements.
• Mnemonic: C for Committee (no order), P for Password (order matters).

Key formulas

• ⁿPr = n!/(n−r)!; ⁿCr = n!/(r!(n−r)!); ⁿPr = r!·ⁿCr.
• Fundamental counting principle: multiply choices of independent stages.
• Arrangements of n objects with repeats p,q,…: n!/(p! q! …).
• Circular permutations of n distinct objects: (n−1)!.
• Selections of at least one from n: 2ⁿ − 1.
• ⁿCr = ⁿC(n−r); Σ ⁿCr = 2ⁿ.

Usage: decide whether order matters (P) or not (C) and whether repetition is allowed.

Worked Example

Problem: How many distinct arrangements can be made using all the letters of the word SUCCESS? In how many of these does the word begin with S?

Solution:
The word SUCCESS has 7 letters: S, U, C, C, E, S, S. The repeated letters are S (3 times) and C (2 times); U and E occur once.

Total arrangements = 7! divided by the factorials of the repeats:
= 7! / (3! · 2!) = 5040 / (6 × 2) = 5040 / 12 = 420.

Arrangements beginning with S: fix one S in the first position. The remaining 6 letters to arrange are S, S, C, C, E, U, in which S occurs twice and C occurs twice:
= 6! / (2! · 2!) = 720 / 4 = 180.

Answer: 420 distinct arrangements in total; 180 of them begin with S.