IF `""^nP_r=840 , "^nC_r=35` then n=

Options

A. 5
B. 6
C. 7
D. none of these

Answer

To solve the problem where \( ^nP_r = 840 \) and \( ^nC_r = 35 \), we will use the formulas for permutations and combinations.

### Step-by-Step Solution:

1. **Write down the formulas for permutations and combinations:**
\[ ^nP_r = \frac{n!}{(n-r)!} \]
\[ ^nC_r = \frac{n!}{r!(n-r)!} \]

2. **Substitute the given values into the formulas:**
From the problem, we have:
\[ \frac{n!}{(n-r)!} = 840 \quad \text{(1)} \]
\[ \frac{n!}{r!(n-r)!} = 35 \quad \text{(2)} \]

3. **Divide equation (1) by equation (2):**
\[ \frac{\frac{n!}{(n-r)!}}{\frac{n!}{r!(n-r)!}} = \frac{840}{35} \]
This simplifies to:
\[ \frac{1}{\frac{1}{r!}} = \frac{840}{35} \]
Thus:
\[ r! = \frac{840}{35} \]

4. **Calculate \( r! \):**
\[ r! = 24 \]
The factorial value of 24 corresponds to:
\[ r = 4 \quad \text{(since \( 4! = 24 \))} \]

5. **Substitute \( r \) back into equation (1):**
Now substitute \( r = 4 \) into equation (1):
\[ \frac{n!}{(n-4)!} = 840 \]

6. **Expand the left-hand side:**
\[ n(n-1)(n-2)(n-3) = 840 \]

7. **Find \( n \) by trial and error or factorization:**
We can express 840 as a product of four consecutive integers:
\[ 840 = 7 \times 6 \times 5 \times 4 \]
This suggests that \( n = 7 \).

8. **Final answer:**
Therefore, the value of \( n \) is:
\[ n = 7 \]