Permutations and Combinations

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The Fundamental Principle of Counting states: If an event can occur in m ways and after it happens in any one of these ways, a second event can occur in n ways, then both events together can occur in

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n! (n factorial) = n × (n-1) × (n-2) × ... × 3 × 2 × 1 By definition: • 1! = 1 • n! = n × (n-1)! for n > 1 • 0! = 1 (by convention) Calculating 5!: 5! = 5 × 4 × 3 × 2 × 1 = 120 Alternatively: 5! =

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Permutation: An arrangement of objects where ORDER MATTERS • Involves selection AND arrangement • Example: Forming 3-digit numbers using digits 1,2,3,4,5 • 245, 452, 542 are different permutations Co

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Formula: Number of permutations of n objects taken all at a time = n! Solution for 6 books: Step 1: Identify n = 6 Step 2: Apply formula Number of arrangements = 6! Step 3: Calculate 6! = 6 × 5 × 4 ×

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ⁿPᵣ = Number of permutations of r objects out of n objects Formulas: • ⁿPᵣ = n(n-1)(n-2)...(n-r+1) • ⁿPᵣ = n!/(n-r)! Calculating ₆P₃: Method 1: ₆P₃ = 6 × 5 × 4 = 120 Method 2: ₆P₃ = 6!/(6-3)! = 6!/3

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Step 1: Identify the problem type • We need to form 3-digit numbers • Order matters (247 ≠ 742) • No repetition allowed • This is a permutation problem Step 2: Identify values • Total digits availabl

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ⁿCᵣ = Number of combinations of r objects out of n objects (selections where order doesn't matter) Formulas: • ⁿCᵣ = ⁿPᵣ/r! = n!/(r!(n-r)!) • ⁿCᵣ = n(n-1)(n-2)...(n-r+1)/r! Calculating ₇C₃: Method 1

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Step 1: Identify problem type • Selecting 3 students from 8 • Order doesn't matter (committee composition is same regardless of selection order) • This is a combination problem Step 2: Identify value