Find n , P_(n+2) ^( 4) = 14P


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Question Number 178543 by Acem last updated on 17/Oct/22

$F i n d n, P_{n + 2}^{4} = 14 P_{n}^{3}$
	Answered by Ar Brandon last updated on 18/Oct/22

	$\overset{n + 2}{} P_{4} = 14 \overset{n}{} P_{3} \Rightarrow \frac{(n + 2)!}{(n - 2)!} = 14 \frac{n!}{(n - 3)!}$ $\Rightarrow \frac{(n + 2) (n + 1)}{n - 2} = 14 \Rightarrow n^{2} - 11 n + 30 = 0$ $\Rightarrow (n - 6) (n - 5) = 0 \Rightarrow n = 5 or n = 6$
	Commented by Acem last updated on 18/Oct/22

	$V e r y g o o d S i r!$
	Commented by Acem last updated on 18/Oct/22
	$As an advice my friend, we must find a condition for n befor i.e. {: ((P_4 ^( n+2) : n+2≥ 4 ⇒ n≥ 2)),((P_3 ^( n) : n≥ 3)) } ⇒ n≥ 3 After that we find n and accept only which belonging to the domain above$
	$A s a n a d v i c e m y f r i e n d, w e m u s t f i n d a c o n d i t i o n$ $f o r n b e f o r i . e .$ $\begin{matrix} P_{4}^{n + 2} : n + 2 ⩾ 4 \Rightarrow n ⩾ 2 \\ P_{3}^{n} : n ⩾ 3 \end{matrix}} \Rightarrow n ⩾ 3$ $A f t e r t h a t w e f i n d n a n d a c c e p t o n l y w h i c h$ $b e l o n g i n g t o t h e d o m a i n a b o v e$
	Commented by Ar Brandon last updated on 18/Oct/22
	You're right!
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