Find 'n' if :
(i) `""^(2n)P_(3)=100 ""^(n)P_(2)`
(ii) `""^(n)P_(4)=20xx ""^(n)P_(2)`

Answer

To solve the problem, we need to find the value of \( n \) based on the two given equations involving permutations. Let's break down each part step by step.

### Part (i): \( (2n)P_3 = 100(n)P_2 \)

1. **Write the formula for permutations**:
\[ nPr = \frac{n!}{(n-r)!} \]
Therefore,
\[ (2n)P_3 = \frac{(2n)!}{(2n-3)!} \quad \text{and} \quad (n)P_2 = \frac{n!}{(n-2)!} \]

2. **Substitute into the equation**:
\[ \frac{(2n)!}{(2n-3)!} = 100 \cdot \frac{n!}{(n-2)!} \]

3. **Simplify the left side**:
\[ (2n)(2n-1)(2n-2) = 100 \cdot \frac{n!}{(n-2)!} \]
The right side simplifies to:
\[ 100 \cdot n(n-1) \]

4. **Set the equations equal**:
\[ (2n)(2n-1)(2n-2) = 100n(n-1) \]

5. **Expand both sides**:
\[ 8n^3 - 12n^2 + 4n = 100n^2 - 100n \]

6. **Rearrange the equation**:
\[ 8n^3 - 112n^2 + 104n = 0 \]

7. **Factor out common terms**:
\[ 8n(n^2 - 14n + 13) = 0 \]

8. **Solve for \( n \)**:
- Set \( 8n = 0 \) gives \( n = 0 \) (not valid in this context).
- Solve the quadratic \( n^2 - 14n + 13 = 0 \) using the quadratic formula:
\[ n = \frac{14 \pm \sqrt{(-14)^2 - 4 \cdot 1 \cdot 13}}{2 \cdot 1} = \frac{14 \pm \sqrt{196 - 52}}{2} = \frac{14 \pm \sqrt{144}}{2} = \frac{14 \pm 12}{2} \]
This gives \( n = 13 \) or \( n = 1 \).

### Part (ii): \( (n)P_4 = 20(n)P_2 \)

1. **Write the formula for permutations**:
\[ (n)P_4 = \frac{n!}{(n-4)!} \quad \text{and} \quad (n)P_2 = \frac{n!}{(n-2)!} \]

2. **Substitute into the equation**:
\[ \frac{n!}{(n-4)!} = 20 \cdot \frac{n!}{(n-2)!} \]

3. **Cancel \( n! \) from both sides** (assuming \( n! \neq 0 \)):
\[ \frac{1}{(n-4)!} = 20 \cdot \frac{1}{(n-2)(n-3)(n-4)!} \]

4. **Multiply both sides by \( (n-4)! \)**:
\[ 1 = 20 \cdot \frac{1}{(n-2)(n-3)} \]

5. **Rearrange the equation**:
\[ (n-2)(n-3) = 20 \]

6. **Expand and rearrange**:
\[ n^2 - 5n + 6 - 20 = 0 \implies n^2 - 5n - 14 = 0 \]

7. **Solve for \( n \)** using the quadratic formula:
\[ n = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot (-14)}}{2 \cdot 1} = \frac{5 \pm \sqrt{25 + 56}}{2} = \frac{5 \pm \sqrt{81}}{2} = \frac{5 \pm 9}{2} \]
This gives \( n = 7 \) or \( n = -2 \) (not valid).

### Final Solutions:
- From part (i): \( n = 13 \) or \( n = 1 \)
- From part (ii): \( n = 7 \)