if np4=360 than find n

P4n=360⇒n!n-4!=360⇒nn-1n-2n-3n-4!n-4!=360⇒nn-1n-2n-3=360⇒n4-6n3+11n2-6n-360=0⇒n4-6n3+11n2-66n+60n-360=0⇒n3n-6+11nn-6+60n-6=0⇒n-6n3+11n+60=0⇒n-6=0 and n3+11n+60=0⇒n=6 and n=-3,32,32Only n= 6 is an positive integer value Thus the value of n is 6

if ⁿp₄=360 than find n

{}^{n}P_{4} = 360 \Rightarrow \frac{n!}{(n - 4)!} = 360 \Rightarrow \frac{n (n - 1) (n - 2) (n - 3) (n - 4)!}{(n - 4)!} = 360 \Rightarrow n (n - 1) (n - 2) (n - 3) = 360 \Rightarrow n^{4} - 6 n^{3} + 11 n^{2} - 6 n - 360 = 0 \Rightarrow n^{4} - 6 n^{3} + 11 n^{2} - 66 n + 60 n - 360 = 0 \Rightarrow n^{3} (n - 6) + 11 n (n - 6) + 60 (n - 6) = 0 \Rightarrow (n - 6) (n^{3} + 11 n + 60) = 0 \Rightarrow (n - 6) = 0 a n d (n^{3} + 11 n + 60) = 0 \Rightarrow n = 6 a n d n = - 3, \frac{3}{2}, \frac{3}{2} O n l y n = 6 i s a n p o s i t i v e i n t e g e r v a l u e . T h u s t h e v a l u e o f n i s 6 .

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