Permutations & Combinations 🎲

Arrangements aur selections — kitne tareekon se kaam kar sakte ho? Yahi counting chapter hai!

🎭 Order Matters ya Nahi?

Imagine karo 3 doston ka group hai: Ali, Bhavna, Chetan. Ek race mein pehle, doosre, teesre kaun hoga? Yahan order matter karta hai — ABC aur BAC alag results hain. Lekin ek committee mein sirf 2 choose karne hain — koi bhi 2. Yahan {Ali,Bhavna} aur {Bhavna,Ali} ek hi cheez hai — order matter nahi karta! Yahi hai Permutation vs Combination ka fark.

1. Fundamental Counting Principle

Agar ek kaam ko m tareekon se kiya ja sake aur doosra kaam n tareekon se, toh dono saath karne ke m × n tareeqe hain. Yeh principle P&C ki neev hai!

Ek ke baad ek kaam: m × n × p × ... tareeqe

Example: Agar 3 shirts hain aur 4 pants hain, toh 3×4 = 12 alag combinations pehne ja sakte hain.

✅ Ek 3-digit PIN mein: First digit 1-9, baaki 0-9. Total PINs kitne?

First digit: 9 choices (1-9, not 0)
Second digit: 10 choices (0-9)
Third digit: 10 choices (0-9)
Total = 9 × 10 × 10 = 900 PINs

2. Factorial — n!

n! (n factorial) = n se lekar 1 tak ke saare natural numbers ka product. Yeh counting ka building block hai.

n! = n × (n−1) × (n−2) × ... × 2 × 1
0! = 1 (yeh definition hai, proof nahi — isliye yaad karo!)

Values: 1!=1, 2!=2, 3!=6, 4!=24, 5!=120, 6!=720, 7!=5040

Shortcut: n! = n × (n−1)!

3. Permutation — Order matters!

n distinct objects mein se r objects select karke arrange karne ke tareeqe — yahan order important hai. P ko "nPr" ya "P(n,r)" likhte hain.

ⁿPᵣ = n! / (n−r)!    (jab n ≥ r)

Intuition: Pehli jagah n choices, doosri jagah (n−1) choices, ... r-vi jagah (n−r+1) choices → n×(n−1)×...×(n−r+1) = n!/(n−r)!

Tree Diagram: {A,B,C} mein se 2 arrange karo (³P₂)

Start A B C B C A C A B AB AC BA BC CA CB Total = 6 ³P₂ = 6 ✓

✅ 8 logon mein se 3 logon ko first, second, third rank deni hai. Kitne tareeqe?

n=8, r=3 → ⁸P₃ = 8!/(8−3)! = 8!/5!
= 8 × 7 × 6 = 336 tareeqe

4. Combination — Order matters nahi!

n distinct objects mein se r objects select karne ke tareeqe (arrange nahi, sirf choose). Yahan {A,B} aur {B,A} same selection hai.

ⁿCᵣ = n! / (r! × (n−r)!)

Relation: ⁿCᵣ = ⁿPᵣ / r!   (permutations ko r! se divide karo kyunki r! arrangements same selection represent karte hain)

🔴 Permutation ⁿPᵣ

Order MATTERS. Har arrangement alag.

AB ≠ BA

Formula: n!/(n−r)!

Use: Races, ranks, passwords

🟢 Combination ⁿCᵣ

Order DOESN'T MATTER. Sirf selection.

{A,B} = {B,A}

Formula: n!/(r!(n−r)!)

Use: Teams, committees, sets

✅ 10 students mein se 3 ki committee banani hai. Kitne tareeqe?

n=10, r=3 → ¹⁰C₃ = 10!/(3! × 7!)
= (10 × 9 × 8)/(3 × 2 × 1) = 720/6 = 120 tareeqe

5. Important Properties of ⁿCᵣ

ⁿCᵣ = ⁿC(n−r)   (complement property)
ⁿC₀ = ⁿCₙ = 1
ⁿCᵣ + ⁿC(r−1) = ⁿ⁺¹Cᵣ   (Pascal's identity)

¹⁰C₇ = ¹⁰C₃ = 120 — complement property se bade r wale easier nikale ja sakte hain!

6. Special Cases

SituationFormulaExample
n objects mein se sabko arrange karon! ways5 log = 5! = 120
Kuch identical objects ke saathn! / (p! × q! × ...)MISSISSIPPI = 11!/(4!4!2!)
Circle mein arrange karo(n−1)!6 log round table = 5! = 120
Atleast ek select karo2ⁿ − 1n items mein se: 2ⁿ−1 non-empty subsets

7. Practice Questions

Q1. ⁷P₃ aur ⁷C₃ nikalo.

Solution

⁷P₃ = 7×6×5 = 210 | ⁷C₃ = 210/6 = 35

Q2. "INDIA" ke letters ko kitne tareeqon se arrange kar sakte hain?

Solution

5 letters mein I do baar → 5!/2! = 120/2 = 60

Q3. 12 teams hain, top 4 choose karne hain. Kitni ways mein selection ho sakta hai?

Solution

¹²C₄ = (12×11×10×9)/(4×3×2×1) = 11880/24 = 495

Q4. ⁿCᵣ = ⁿC(r+1), agar n=11 aur r=5 to check karo.

Solution

¹¹C₅ = 462, ¹¹C₆ = 462 — haan, complement property se equal hain! (r + (r+1) = n → 5+6=11 ✓)

Q5. 6 logon mein se 4 logon ki team banani hai jisme ek specific person hamesha included ho. Kitne tareeqe?

Solution

Woh 1 person fix hai. Baaki 5 mein se 3 choose karo: ⁵C₃ = 10 tareeqe

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