Learning what a factorial is will help you work out these problems. For example, 5! is the product of the integer 5 and every integer less than it, stopping at the integer 1. 5! = 5 * 4 * 3 * 2 * 1. For these problems, even if we replace the number 5 with a variable n, the factorial still holds the same property.

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

What this means is we can also rewrite 5! as 5 * 4!. With a variable n, this looks like n! = n * (n-1)!

Hope this helps!

a)

(n+1)!/n! = 9

(n+1)! = (n+1) * n!

(n+1)*n!/n! = 9

(n+1) = 9

n=8

b)

n!/(n-2)! = 20

n! = n * (n-1) * (n-2)!

n*(n-1)*(n-2)! / (n-2)! = 20

n*(n-1) = 20

n = 5

c)

3(n+1)! / (n-1)! = 126

3*(n+1)*(n)*(n-1)! / (n-1)! = 126

3*(n+1)*(n) = 126

(n+1)*(n) = 42

n = 6

d)

2n! / (n-3)! = 84n

2* n * (n-1) * (n-2) * (n-3)! / (n-3)! = 84n

2 * n * (n-1) * (n-2) = 84n

2 * (n-1) * (n-2) = 84

(n-1) * (n-2) = 42

n = 8