Question

Find n in this equations: 9P(n,5)=P(n,3).P(9,3)​

Answer 1

Explanation:

9P(n,5)=P(n,3).P(9,3)

9n(n-1)(n-2)(n-3)(n-4)=n(n-1)(n-2)×9×8×7

(n-3)(n-4)=8×7

(n²-4n-3n+12)=56

(n²-7n+12)=56

n²-7n+12-56=0

n²-7n-44=0

n²-11n+4n-44=0

n(n-11)+4(n-11)=0

(n-11)(n+4)=0

n=11,-4

n=-4(rejected as n is a natural number.)

Hence n=11

Answer 2

`\begin{aligned}&9P(n,5)=P(n,3)\cdot P(9,3)\\&9\cdot(n!)/((n-5)!)=(n!)/((n-3)!)\cdot(9!)/(6!)\\&(n!)/((n-5)!)=(n!)/((n-3)!)\cdot7\cdot8\\&(1)/((n-5)!)=(56)/((n-3)!)\\&(n-4)(n-3)=56\\&n^2-3n-4n+12-56=0\\&n^2-7n-44=0\\&n^2-11n+4n-44=0\\&n(n-11)+4(n-11)=0\\&(n+4)(n-11)=0\\&n=-4 \vee n=11\end`

`n` can't be negative, therefore
`n=11`