Combinations

• nCr = n!/(r!(n−r)!) — choosing r items from n without regard to order.
• Pascal's identity: nCr = n−1Cr−1 + n−1Cr.
• Pascal's triangle and its patterns.
• Sum identities: ΣnCr = 2ⁿ, symmetry nCr = nCn−r.
• nCr + nCr+1 = n+1Cr+1 (Pascal's rule) and applications.

Level 1 — Definition and basic properties

Combinations: nCr = nPr / r! = n!/(r!(n−r)!). Order does NOT matter. Think: choose r from n.

Key values: nC0 = 1, nC1 = n, nCn = 1, nCr = nCn−r (symmetry).

Relation to permutations: nPr = nCr × r! (select r, then arrange them).

Symmetry (complementary selection): nCr = nCn−r. Choosing r to include = choosing (n−r) to exclude. Use when r > n/2 to compute a smaller equivalent nCn−r instead.

nCr + nCr+1 = n+1Cr+1 (Pascal's rule): Proof — right side = (n+1)!/((r+1)!(n−r)!) = n!/r!(n−r)! + n!/(r+1)!(n−r−1)! = nCr + nCr+1. □

Level 2 — Pascal's triangle, sum identities, advanced patterns

Pascal's triangle: Row n has entries nC0, nC1, …, nCn. Each entry = sum of two above it (Pascal's rule). Row sums: Σ(k=0 to n) nCk = 2ⁿ.

Sum identities:

• nC0 + nC1 + nC2 + … + nCn = 2ⁿ (sum of all selections from n items = 2ⁿ subsets).
• nC0 − nC1 + nC2 − … = 0 (alternating sum = 0, from (1−1)ⁿ = 0).
• nC0 + nC2 + nC4 + … = 2ⁿ⁻¹ (even-indexed = 2ⁿ⁻¹).
• nC1 + nC3 + nC5 + … = 2ⁿ⁻¹ (odd-indexed = 2ⁿ⁻¹).

Vandermonde's identity: Σ(k=0 to r) mCk × nCr−k = m+nCr. Special case: Σ(nCk)² = 2nCn.

Maximum value of nCr: Maximum at r = n/2 (if n even) or r = (n±1)/2 (if n odd). nCr is unimodal — increases to middle then decreases.

Combinatorial interpretation: nC2 = number of line segments connecting n points = n(n−1)/2. nC3 = number of triangles from n points.

JEE pattern — choosing with restrictions:

• At least one from group A: use complementary — total − (none from A).
• Exactly k from group A: AC_k × BC(r−k) where B is rest of group.

NCERT spotlight

Distinguish nPr and nCr by asking: "Does order matter?" — committees, selections, subsets → nCr; arrangements, words, ranks → nPr. Pascal's rule is the basis for Binomial theorem (Class 11, next chapter). The identity nCr = nCn−r is useful to simplify: 12C9 = 12C3 = 220 (much easier computation).

In how many ways can a committee of 3 men and 2 women be formed from 6 men and 5 women?

Step 1 — Choose 3 men from 6: 6C3 = 6!/(3!×3!) = 720/36 = 20.
Step 2 — Choose 2 women from 5: 5C2 = 5!/(2!×3!) = 120/12 = 10.
Step 3 — Total committees = 20 × 10 = 200. (Multiplication rule — independent choices.)

Prove Pascal's identity: nCr = n−1Cr−1 + n−1Cr.

Step 1 — Combinatorial proof: Consider n objects, fix one object O.
Step 2 — Case 1: O is included in the r-selection. Choose remaining r−1 from the other n−1 objects: n−1Cr−1 ways.
Step 3 — Case 2: O is excluded. Choose all r from the remaining n−1 objects: n−1Cr ways.
Step 4 — These cases are mutually exclusive and exhaustive, so nCr = n−1Cr−1 + n−1Cr. □
Step 5 — Algebraic verification: n−1Cr−1 + n−1Cr = (n−1)!/((r−1)!(n−r)!) + (n−1)!/(r!(n−r−1)!)
         = (n−1)![r + (n−r)] / (r!(n−r)!) = n·(n−1)! / (r!(n−r)!) = n!/(r!(n−r)!) = nCr. ✓

1. Evaluate 10C4.
2. In how many ways can 4 cards be chosen from a standard deck of 52?
3. Find n if nC2 = 45.
4. Prove that nC0 + nC1 + … + nCn = 2ⁿ using the binomial theorem with x = y = 1.
5. Stretch: Prove Vandermonde's identity Σ(k=0 to r) mCk × nCr−k = m+nCr using a combinatorial argument (choosing r people from m men and n women).

Open the Practice tab for graded questions on Permutations & Combinations — Combinations.

• Use the platform-native live simulation or PhET-style tool for this topic (number line, Venn, physics playground, molecule builder, sensor dashboard, etc.).
• Mirror / body / home activity: physically do the concept (count objects, measure, role-play) and photograph or describe for portfolio.
• Voice or text reflection with AI Mentor: explain the concept to a younger student or family member.

• "Explain this concept to a Class 6 student using one real example from an Indian home, school, market, or festival."
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• Stretch: "How does this connect to coding, robotics, money, health, environment, or a future career?"

• Complete the core practice + one extension activity (photo, table, short reflection, or mini-project) for base XP + topic badge.
• 5-7 day streak or family discussion note = multiplier + visible artifact in parent/principal dashboard.
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• One hands-on project or measurement using the Drishti kit or household items that makes the concept physical.
• Direct link to at least one Future Skill track (Money Management, Green Tech, Cyber Defenders, Micro-Entrepreneurship, AI Mastery, Sustainable Living, Personality Development).
• Coding extension where relevant (simple script, simulation, or data logging).

This material emphasises experiential "learning by doing", competency (apply/create/analyse), vocational exposure, critical thinking, and multidisciplinary connections. Designed to feed live worlds, AI Mentor (with memory), gamification, robotics, parent analytics, and future skills — not just exam prep.

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Open the Practice tab for aligned questions (easy/medium/hard + case-based) with full AI scaffolding.

See curriculum for cross-links and the full future-skills/robotics chapters.