If `.^(n)P_(r) = 720` and `.^(n)C_(r) = 120`, then find r.
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Answer

To solve the problem, we need to find the value of \( r \) given the equations \( nP_r = 720 \) and \( nC_r = 120 \).

### Step-by-step Solution:

1. **Understanding the Formulas**:
- The formula for permutations is given by:
\[ nP_r = \frac{n!}{(n-r)!} \]
- The formula for combinations is given by:
\[ nC_r = \frac{n!}{r!(n-r)!} \]

2. **Setting up the Equations**:
- From the problem, we have:
\[ nP_r = 720 \quad \text{(1)} \]
\[ nC_r = 120 \quad \text{(2)} \]

3. **Expressing \( nC_r \) in terms of \( nP_r \)**:
- We can express \( nC_r \) using \( nP_r \):
\[ nC_r = \frac{nP_r}{r!} \]
- Substituting equation (1) into this gives:
\[ nC_r = \frac{720}{r!} \]

4. **Setting up the equation from \( nC_r \)**:
- From equation (2), we have:
\[ \frac{720}{r!} = 120 \]

5. **Solving for \( r! \)**:
- Rearranging the equation gives:
\[ 720 = 120 \cdot r! \]
- Dividing both sides by 120:
\[ r! = \frac{720}{120} = 6 \]

6. **Finding \( r \)**:
- Now we need to find the value of \( r \) such that \( r! = 6 \).
- The factorial values are:
- \( 1! = 1 \)
- \( 2! = 2 \)
- \( 3! = 6 \)
- Therefore, \( r = 3 \).

### Final Answer:
The value of \( r \) is \( 3 \).

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