If `.^(n)P_(r)`= 2520 and `.^(n)C_(r)= 21`, then what is the value of `.^(n+1)C_(R+1)` ?

Options

A. 7
B. 14
C. 28
D. 56

Answer

To solve the problem step by step, we will use the formulas for permutations and combinations.

### Step 1: Understand the given information
We have:
- \( nP_r = 2520 \)
- \( nC_r = 21 \)

### Step 2: Write the formulas for permutations and combinations
The formulas are:
- \( nP_r = \frac{n!}{(n-r)!} \)
- \( nC_r = \frac{n!}{r!(n-r)!} \)

### Step 3: Relate the two equations
From the combination formula, we can express \( n! \) in terms of \( nC_r \) and \( r! \):
\[ nC_r = \frac{n!}{r!(n-r)!} \implies n! = nC_r \cdot r! \cdot (n-r)! \]

### Step 4: Substitute \( nP_r \) into the combination formula
We know that:
\[ nP_r = nC_r \cdot r! \implies 2520 = 21 \cdot r! \]

### Step 5: Solve for \( r! \)
From the equation above:
\[ r! = \frac{2520}{21} = 120 \]

### Step 6: Determine the value of \( r \)
Now we find \( r \) such that \( r! = 120 \). We know:
\[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \]
Thus, \( r = 5 \).

### Step 7: Substitute \( r \) back into the permutation equation to find \( n \)
Now we substitute \( r \) back into the permutation formula:
\[ nP_r = \frac{n!}{(n-r)!} = 2520 \]
This can be rewritten as:
\[ \frac{n!}{(n-5)!} = 2520 \]
This means:
\[ n(n-1)(n-2)(n-3)(n-4) = 2520 \]

### Step 8: Find \( n \) through trial and error
We will try different values of \( n \):
- For \( n = 7 \):
\[ 7 \times 6 \times 5 \times 4 \times 3 = 2520 \]
This works, so \( n = 7 \).

### Step 9: Calculate \( n+1 \) and \( r+1 \)
Now we need to find \( n + 1 \) and \( r + 1 \):
- \( n + 1 = 8 \)
- \( r + 1 = 6 \)

### Step 10: Calculate \( n+1C_{r+1} \)
Now we calculate:
\[ n+1C_{r+1} = 8C_6 = \frac{8!}{6! \cdot 2!} \]
Calculating this gives:
\[ = \frac{8 \times 7}{2 \times 1} = \frac{56}{2} = 28 \]

### Final Answer
Thus, the value of \( n+1C_{r+1} \) is \( 28 \).

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