Combinations — Teams & Committees
Combinations — team और committee चुनना
title
Combinations — Teams & Committees
- Permutation & Combination
- Combinations — Teams & Committees
Hello दोस्तों! MeraExam की एक और class में आपका स्वागत है। आज का topic है — Combinations — team और committee चुनना। बिलकुल zero से, एकदम आसान भाषा में। चलिए शुरू करते हैं!
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Learning Objective
Count selections with nCr, use nCr = nC(n−r) to shorten work, and solve team/committee questions.
🎯 Learning Objective
Count selections with nCr, use nCr = nC(n−r) to shorten work, and solve team/committee questions.
💡 Concept
- nCr = n!/[r!(n−r)!] counts SELECTIONS — order does not matter
- Team, committee, handshake → combination; race positions, photos in a row → permutation
- Symmetry rule: nCr = nC(n−r), so 15C11 becomes the easy 15C4
- nC0 = nCn = 1 and nC1 = n
- Quick calculation: nCr = (r falling terms from n)/r!, e.g. 7C3 = (7×6×5)/(3×2×1) = 35
🧮 Key Formulas
nCr = n!/[r!(n−r)!]
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nCr = nC(n−r)
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Handshakes among n people = nC2 = n(n−1)/2
✏️ Easy Example
Q. In how many ways can 2 students be chosen from 5 for a quiz team?
- 5C2 = (5 × 4)/(2 × 1)
- = 20 ÷ 2
Answer: 10
🇮🇳 Real-Life Example
A mohalla WhatsApp group of 10 members: total possible one-to-one chats = 10C2 = 45 — every pair once, order irrelevant.
📝 Exam-Level Example
Q. In how many ways can a cricket team of 11 be selected from 15 players?
- 15C11 = 15C4
- = (15×14×13×12)/(4×3×2×1)
- = 32760 ÷ 24
Answer: 1365
📝 Exam-Level Example
Q. A committee of 3 must have 2 men from 4 men and 1 woman from 3 women. How many committees are possible?
- Men: 4C2 = 6
- Women: 3C1 = 3
- Total = 6 × 3
Answer: 18
🪄 Memory Trick
Big r? Flip it: nCr = nC(n−r). 15C11 → 15C4 turns a page of factorials into four small numbers.
⚠️ Common Mistakes
- ❌ Using permutation for team selection — selection never cares about order
- ❌ Multiplying alternative cases — 'this OR that' means add the counts
🏆 Exam Tips
- ✅ 'AND' between selections → multiply; 'OR' / 'at least' → add the cases
- ✅ Handshakes, matches, diagonals-type questions all reduce to nC2
📌 Summary
- nCr = selections, order ignored
- nCr = nC(n−r) — always flip a big r
- Compound committees: multiply group-wise choices
- Pairs among n = nC2 = n(n−1)/2