Pascal's triangle & nCr

Answer

Start and end every row with 1; each inside number is the sum of the two directly above it. Rows: 1 / 1 1 / 1 2 1 / 1 3 3 1.

Answer

The coefficients of (a + b)ⁿ. E.g. row 3 (1, 3, 3, 1) → (a + b)³ = a³ + 3a²b + 3ab² + b³.

Answer

Take r = 3 factors counting down from 8 on top, r! on the bottom: ⁸C₃ = (8×7×6)/(3×2×1) = 56. The big factorials cancel — never expand them in full. (ⁿCᵣ = n!/(r!(n − r)!).)

Answer

5!/(2! 3!) = (5 × 4)/(2 × 1) = 10.

Answer

Type n, then MATH → ▶ (PRB) → 3: nCr, then r, then ENTER. E.g. 10 nCr 4 = 210.

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Small n (about ≤ 6) and you want the WHOLE expansion → Pascal's triangle is fastest. Large n, or you only need ONE term/coefficient → use ⁿCᵣ (the general term ⁿCᵣaⁿ⁻ʳbʳ) and skip the rest.

Answer

n + 1 terms. E.g. (x + 2)⁹ has 10 terms.

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The power of a falls from n to 0; the power of b rises from 0 to n; in every term the two powers sum to n.

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Both equal 1 (the first and last coefficient of every row). The row is symmetric: ⁿCᵣ = ⁿCₙ₋ᵣ.

Answer

ⁿCᵣ is the entry in row n, position r (counting from 0). Row 5 = ⁵C₀, ⁵C₁, …, ⁵C₅ = 1, 5, 10, 10, 5, 1.