Sum
Find n if: `("n"!)/(3!("n" - 3)!) : ("n"!)/(5!("n" - 5)!)` = 5:3
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Solution
`("n"!)/(3!("n" - 3)!) : ("n"!)/(5!("n" - 5)!)` = 5:3
∴ `("n"!)/(3!("n" - 3)!)xx(5!("n"-5)!)/("n"!) = 5/3`
`∴("n"!)/(3!("n"-3)("n"-4)("n"-5)!) xx(5xx4xx3!("n"-5)!)/"n!"=5/3`
∴ `(5xx4)/(("n" - 3)("n" - 4)) = 5/3`
∴ (n – 3) (n – 4) = `(20 xx 3)/(5)`
∴ (n – 3) (n – 4) = 12
∴ (n – 3) (n – 4) = 4 × 3
Comparing on both sides, we get
∴ n – 3 = 4
∴ n = 7
Concept: Concept of Factorial Function
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