If `""^(n)P_r=720 ""^(n)C_(r)` then r is equal to

Options

A. 3
B. 7
C. 6
D. 4

Answer

To solve the problem where \( nP_r = 720 \) and \( nC_r \), we can follow these steps:

### 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 equation
According to the problem, we have:
\[ nP_r = 720 \cdot nC_r \]
Substituting the formulas, we get:
\[ \frac{n!}{(n-r)!} = 720 \cdot \frac{n!}{r!(n-r)!} \]

### Step 3: Simplify the equation
We can cancel \( n! \) and \( (n-r)! \) from both sides (assuming \( n! \neq 0 \) and \( (n-r)! \neq 0 \)):
\[ 1 = 720 \cdot \frac{1}{r!} \]
This simplifies to:
\[ r! = 720 \]

### Step 4: Find the value of \( r \)
Now we need to find \( r \) such that \( r! = 720 \). We can calculate factorials:
- \( 1! = 1 \)
- \( 2! = 2 \)
- \( 3! = 6 \)
- \( 4! = 24 \)
- \( 5! = 120 \)
- \( 6! = 720 \)

Thus, we find that \( r = 6 \).

### Conclusion
The value of \( r \) is \( 6 \).

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