Find the value of n if:
(i) `(n+2)! = 12n!`
(ii) `(n+2)! = 60(n-1)!`
(iii) `(n+3)! = 2550 (n+1)!`
(iv) `(n-2)! = 132. (n-4)!`.

Answer

Let's solve the equations step by step.

### (i) Solve for \( n \) in the equation \( (n+2)! = 12n! \)

1. Rewrite the left side using the factorial property:
\[ (n+2)! = (n+2)(n+1)n! \]
So, the equation becomes:
\[ (n+2)(n+1)n! = 12n! \]

2. Cancel \( n! \) from both sides (assuming \( n! \neq 0 \)):
\[ (n+2)(n+1) = 12 \]

3. Expand the left side:
\[ n^2 + 3n + 2 = 12 \]

4. Rearrange to form a quadratic equation:
\[ n^2 + 3n - 10 = 0 \]

5. Factor the quadratic:
\[ (n + 5)(n - 2) = 0 \]

6. Set each factor to zero:
\[ n + 5 = 0 \quad \text{or} \quad n - 2 = 0 \]
This gives:
\[ n = -5 \quad \text{or} \quad n = 2 \]

7. Since factorials of negative numbers are not defined, we take:
\[ n = 2 \]

### (ii) Solve for \( n \) in the equation \( (n+2)! = 60(n-1)! \)

1. Rewrite the left side:
\[ (n+2)! = (n+2)(n+1)n! \]
The equation becomes:
\[ (n+2)(n+1)n! = 60(n-1)! \]

2. Cancel \( (n-1)! \) from both sides:
\[ (n+2)(n+1)n = 60 \]

3. Rearrange:
\[ n^3 + 3n^2 + 2n - 60 = 0 \]

4. Use trial and error to find a root. Testing \( n = 3 \):
\[ 3^3 + 3(3^2) + 2(3) - 60 = 27 + 27 + 6 - 60 = 0 \]
So, \( n = 3 \) is a root.

5. Factor out \( (n - 3) \) using synthetic division:
\[ n^3 + 3n^2 + 2n - 60 = (n - 3)(n^2 + 6n + 20) \]

6. The quadratic \( n^2 + 6n + 20 \) has no real roots (discriminant \( 36 - 80 < 0 \)), so:
\[ n = 3 \]

### (iii) Solve for \( n \) in the equation \( (n+3)! = 2550(n+1)! \)

1. Rewrite the left side:
\[ (n+3)! = (n+3)(n+2)(n+1)n! \]
The equation becomes:
\[ (n+3)(n+2)(n+1)n! = 2550(n+1)! \]

2. Cancel \( (n+1)! \):
\[ (n+3)(n+2)n = 2550 \]

3. Rearrange:
\[ n^3 + 5n^2 + 6n - 2550 = 0 \]

4. Use trial and error to find a root. Testing \( n = 10 \):
\[ 10^3 + 5(10^2) + 6(10) - 2550 = 1000 + 500 + 60 - 2550 = 0 \]
So, \( n = 10 \) is a root.

5. Factor out \( (n - 10) \) using synthetic division:
\[ n^3 + 5n^2 + 6n - 2550 = (n - 10)(n^2 + 15n + 255) \]

6. The quadratic \( n^2 + 15n + 255 \) has no real roots (discriminant \( 225 - 1020 < 0 \)), so:
\[ n = 10 \]

### (iv) Solve for \( n \) in the equation \( (n-2)! = 132(n-4)! \)

1. Rewrite the left side:
\[ (n-2)! = (n-2)(n-3)(n-4)! \]
The equation becomes:
\[ (n-2)(n-3)(n-4)! = 132(n-4)! \]

2. Cancel \( (n-4)! \):
\[ (n-2)(n-3) = 132 \]

3. Rearrange:
\[ n^2 - 5n + 6 - 132 = 0 \]
\[ n^2 - 5n - 126 = 0 \]

4. Factor the quadratic:
\[ (n - 14)(n + 9) = 0 \]

5. Set each factor to zero:
\[ n - 14 = 0 \quad \text{or} \quad n + 9 = 0 \]
This gives:
\[ n = 14 \quad \text{or} \quad n = -9 \]

6. Since factorials of negative numbers are not defined, we take:
\[ n = 14 \]

### Summary of Solutions:
1. \( n = 2 \)
2. \( n = 3 \)
3. \( n = 10 \)
4. \( n = 14 \)