If `.^(n)P_(r)=720 and .^(n)C_(r)=120` then find the value of r.

Options

A. 3
B. 5
C. 4
D. 7

Explanation

To solve the problem, we need to find the value of \( r \) given the equations for permutations and combinations:

1. \( ^nP_r = 720 \)
2. \( ^nC_r = 120 \)

### Step 1: Write the formulas for permutations and combinations
The formulas for permutations and combinations are:
- \( ^nP_r = \frac{n!}{(n-r)!} \)
- \( ^nC_r = \frac{n!}{r!(n-r)!} \)

### Step 2: Set up the equations
From the given information, we can set up the following equations:
1. \( \frac{n!}{(n-r)!} = 720 \) (Equation 1)
2. \( \frac{n!}{r!(n-r)!} = 120 \) (Equation 2)

### Step 3: Divide Equation 1 by Equation 2
To eliminate \( n! \) from both equations, we divide Equation 1 by Equation 2:
\[ \frac{\frac{n!}{(n-r)!}}{\frac{n!}{r!(n-r)!}} = \frac{720}{120} \]

### Step 4: Simplify the equation
This simplifies to:
\[ \frac{1}{\frac{1}{r!}} = \frac{720}{120} \]
\[ r! = \frac{720}{120} \]

### Step 5: Calculate the right side
Calculating \( \frac{720}{120} \):
\[ \frac{720}{120} = 6 \]
So we have:
\[ r! = 6 \]

### Step 6: Determine the value of \( r \)
The value of \( r \) such that \( r! = 6 \) is:
\[ r = 3 \quad \text{(since \( 3! = 6 \))} \]

### Final Answer
Thus, the value of \( r \) is:
\[ \boxed{3} \]