hitunglah n dari persamaan berikut! a. (n + 2)! / (n + 1)! = 5 b. (n + 2)! / n! = 56 c. (n - 3 )! / (n - 4)! = 11 d. n! / (n - 2)! = 20 tolong dibantu ya tolong

Rom A
[tex]\frac{(n+2)!}{(n+1)!} = 5 \\ \frac{(n+2)(n+1)!}{(n+1)!}= 5 \\ n+2 = 5 \\ \therefore n=3[/tex]
Rom B
[tex]\frac{(n+2)!}{n!}=56 \\ \frac{(n+2)(n+1)n!}{n!}=56 \\ (n+2)(n+1)=56 \\ (n+2)(n+1)=8 \times 7 \\ \therefore n = 6[/tex]
Rom C
[tex]\frac{(n-3)!}{(n-4)!} = 11 \\ \frac{(n-3)(n-4)!}{(n-4)!} = 11 \\ n-3=11 \\ \therefore n = 14 [/tex]
Rom D
[tex]\frac{n!}{(n-2)!} = 20 \\ \frac{n(n-1)(n-2)!}{(n-2)!} = 20 \\ n(n-1) = 20 \\ n(n-1) = 5 \times 4 \\ \therefore n = 5 [/tex]