If 4×ⁿC₅ = 9 ×^n-1C₅ then the value of n will be?

Question

Question

Answer 1

Formula used:

\(^nC_r=\frac{n!}{r!(n-r)!}\)

n! = n × (n - 1) × (n - 2).......3 × 2 × 1

Calculation:

Given that,

4 × nC5 = 9 × n-1C5

Using the above formula

\(4×\frac{n!}{5!(n-5)!}=9×\frac{(n-1)!}{5![(n-1)-5]!}\)

⇒ \(4×\frac{n× (n-1)!}{5!(n-5)(n-6)!}=9×\frac{(n-1)!}{5![(n-6]!}\)

⇒ \(\frac{4n}{(n-5)} = 9\)

⇒ 9n - 45 = 49

⇒ 5n = 4n

⇒ n = 9