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Permutation & Combination

The Chapter That Decides Ranks

Here's a fact that should wake you up: in almost every advanced exam — SSC CGL Tier-II, IBPS PO Mains, CAT, UPSC CSAT — at least one question appears where the entire solution is a single line like ¹⁰C₃ = 120 or 8! = 40,320. Students who understand the *logic* finish it in 20 seconds. Students who memorised formulas without understanding spend 4 minutes and still pick the wrong option.

Why? Because Permutation & Combination (P&C) is not really a "maths" chapter. It is a counting chapter. And counting is something you've done since childhood — you just never learned its two secret rules.

Learning goals for this chapter:

1. Master the Fundamental Principle of Counting (AND rule vs OR rule).
2. Clearly distinguish Permutation (order matters) from Combination (order doesn't).
3. Handle circular arrangements, repetitions, and identical objects.
4. Solve advanced patterns: selection with constraints, distribution, derangements, and grouping.
5. Spot the examiner's traps before they trap you.

1. The Fundamental Principle of Counting — The Mother of All P&C

Imagine you have 3 shirts and 2 trousers. How many outfits? You don't need a formula — you pair each shirt with each trouser: 3 × 2 = 6.

That's it. That's the Multiplication Rule: if task A can be done in *m* ways AND task B in *n* ways, both together can be done in m × n ways.

Now the twin rule: if you must do task A OR task B (only one of them), the total is m + n ways.

> The golden question to ask yourself: "Am I doing both tasks, or choosing one?"
> - Both → AND → Multiply (×)
> - One of them → OR → Add (+)

🧠 Active Recall: A menu has 4 soups, 6 main courses, and 3 desserts. How many complete meals (one of each)? And how many choices if you pick *either* a soup *or* a dessert only? *(Answer: 4×6×3 = 72; and 4+3 = 7.)*

2. Factorial — The Engine of P&C

n! = n × (n−1) × (n−2) × … × 2 × 1, and by definition 0! = 1.

Why does 0! = 1? Because there is exactly one way to arrange zero objects — do nothing. This tiny fact saves you in dozens of formula cancellations.

Memorise these cold: 0!=1, 1!=1, 2!=2, 3!=6, 4!=24, 5!=120, 6!=720, 7!=5040, 8!=40320, 9!=362880, 10!=3628800.

🧠 Active Recall: Without a calculator, what is 8! ÷ 6!? *(Answer: 8×7 = 56 — factorials cancel from the bottom up.)*

3. Permutation — When ORDER Matters

A permutation is an arrangement. If you award Gold, Silver, Bronze among 8 athletes, then "A first, B second" is *different* from "B first, A second". Order matters.

Number of permutations of n distinct objects taken r at a time:

$$^nP_r = \frac{n!}{(n-r)!}$$

Special case: arranging all n objects = n!.

With repetition allowed: n^r (e.g., 3-digit codes from 5 digits, digits repeatable = 5³).

With identical objects: if among n objects, p are alike of one kind, q of another:
$$\frac{n!}{p!\,q!\,r!\dots}$$
Classic example: the word ALLAHABAD has 9 letters with A×4, L×2 → arrangements = 9!/(4!·2!) = 7560.

🧠 Active Recall: How many words (meaningful or not) can be formed from the letters of "LEVEL"? *(Answer: 5!/(2!·2!) = 30, since L and E each repeat twice.)*

4. Combination — When ORDER Does NOT Matter

A combination is a selection. Choosing a committee of 3 from 10 people — the committee {Ram, Shyam, Mohan} is the *same* as {Mohan, Ram, Shyam}. Order is irrelevant.

$$^nC_r = \frac{n!}{r!\,(n-r)!}$$

The bridge between the two: ⁿPᵣ = ⁿCᵣ × r! — every selection of r objects can be arranged in r! ways. So a combination is just a permutation with the internal ordering *divided out*.

Key identities (examiners love these):

• ⁿCᵣ = ⁿCₙ₋ᵣ (choosing 3 to include = choosing n−3 to exclude)
• ⁿC₀ = ⁿCₙ = 1
• ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ (Pascal's rule)
• ⁿC₀ + ⁿC₁ + … + ⁿCₙ = 2ⁿ (total subsets of an n-set)

🧠 Active Recall: ¹²C₉ = ? Don't compute — use the identity. *(Answer: ¹²C₃ = 220.)*

5. 🎵 The Mnemonic That Ends All Confusion

> "P for Position, C for Committee."
>
> Permutation → Position, Podium, Password → order matters → P is Picky.
> Combination → Committee, Combo meal, Collection → order doesn't matter → C is Chill.

One more for the counting rules: "AND you multiply, OR you add" — *"And-Multiply, Or-Add"* — say it like a chant: AM-OA.

6. Circular Permutations — The Round Table Trap

Arranging n people in a row: n! ways.
Arranging n people around a circle: (n−1)! ways.

Why? In a circle, rotating everyone by one seat doesn't create a new arrangement — only *relative* positions matter. So we fix one person and arrange the remaining (n−1).

Garlands / necklaces (flipping allowed): (n−1)! / 2, because a necklace looks the same when flipped.

🧠 Active Recall: 6 people around a round table; 6 beads in a necklace. *(Answers: 5! = 120; 5!/2 = 60.)*

7. Advanced Patterns Every Graduation-Level Exam Tests

(a) "Together / Never together": Treat the glued pair as one unit, arrange, then multiply by internal arrangements. "Never together" = Total − Together.

(b) Vowels together / consonants fixed: Same glue technique on letter groups.

(c) At least / At most selections: Break into cases and ADD. E.g., committee of 4 from 5 men, 4 women with *at least 2 women* = (²W + ³W + ⁴W cases) summed.

(d) Distributing n distinct objects into r distinct boxes (no box empty): use inclusion–exclusion or Stirling logic; if boxes identical, it's a partition problem.

(e) Derangements (no object in its original place):
$$D_n = n!\left(1 - \frac{1}{1!} + \frac{1}{2!} - \dots + (-1)^n\frac{1}{n!}\right)$$
Memorise: D₂=1, D₃=2, D₄=9, D₅=44. (The classic "letters in wrong envelopes" question.)

(f) Number of ways to select r from n with repetition allowed: ⁿ⁺ʳ⁻¹Cᵣ (stars and bars — e.g., distributing identical chocolates among children).

(g) Geometry links: Number of lines from n points (no 3 collinear) = ⁿC₂; triangles = ⁿC₃; diagonals of an n-gon = n(n−3)/2.

🧠 Active Recall: In how many ways can 4 letters go into 4 wrong envelopes (none correct)? *(Answer: D₄ = 9.)*

8. The Examiner's Three Favourite Traps

1. The order trap: "Committee of 3 with a president, secretary, treasurer" — sounds like combination, but the posts make it a permutation (or ⁿC₃ × 3!).
2. The "at least one" trap: Students compute one case and forget the rest. Complement method is often faster: Total − (none).
3. The identical-objects trap: Words like BANANA, MISSISSIPPI — forgetting to divide by repeated-letter factorials inflates the answer.

📋 Summary Table

🔁 Review Reminder

Review this chapter after 2 days. Redo the Active Recall prompts without looking, re-derive ⁿPᵣ = ⁿCᵣ × r! once by hand, and re-solve one "at least one" problem using the complement method. Spaced repetition is what converts this chapter from "confusing" to "free marks".