If `.^nP_r=840, .^nC_r=35,` then find `n=`

Options

A. 1
B. 3
C. 5
D. 7

Answer

To solve the problem where \( nP_r = 840 \) and \( nC_r = 35 \), we can follow these steps:

### Step 1: Write down the formulas for permutations and combinations.
The formula for permutations is:
\[ nP_r = \frac{n!}{(n-r)!} \]
The formula for combinations is:
\[ nC_r = \frac{n!}{r!(n-r)!} \]

### Step 2: Substitute the given values into the formulas.
From the problem, we have:
\[ nP_r = 840 \quad \text{and} \quad nC_r = 35 \]
Using the permutation formula:
\[ \frac{n!}{(n-r)!} = 840 \quad \text{(1)} \]
Using the combination formula:
\[ \frac{n!}{r!(n-r)!} = 35 \quad \text{(2)} \]

### Step 3: Relate the two equations.
From equation (2), we can express \( n! \) in terms of \( nP_r \):
\[ n! = 35 \cdot r! \cdot (n-r)! \]
Substituting this into equation (1):
\[ \frac{35 \cdot r! \cdot (n-r)!}{(n-r)!} = 840 \]
This simplifies to:
\[ 35 \cdot r! = 840 \]

### Step 4: Solve for \( r! \).
Dividing both sides by 35:
\[ r! = \frac{840}{35} = 24 \]

### Step 5: Determine the value of \( r \).
Since \( r! = 24 \), we can find \( r \):
\[ r! = 4! = 24 \quad \Rightarrow \quad r = 4 \]

### Step 6: Substitute \( r \) back into equation (1).
Now substitute \( r = 4 \) back into equation (1):
\[ \frac{n!}{(n-4)!} = 840 \]
This can be rewritten as:
\[ n(n-1)(n-2)(n-3) = 840 \quad \text{(3)} \]

### Step 7: Factor 840 to find \( n \).
We can factor 840:
\[ 840 = 7 \times 6 \times 5 \times 4 \]
Thus, we can equate:
\[ n(n-1)(n-2)(n-3) = 7 \times 6 \times 5 \times 4 \]

### Step 8: Solve for \( n \).
By comparing the factors, we can see that:
\[ n = 7 \]

### Conclusion
Thus, the value of \( n \) is:
\[ \boxed{7} \]