If `""^(n)P_r = ""^(n)P_(r+1) and ""^(n)C_r = ""^(n)C_(r-1) `,then the values of n and r are

Options

A. 4,3
B. 3,2
C. 4,2
D. None of these

Answer

To solve the problem, we need to find the values of \( n \) and \( r \) given the equations:

1. \( {}^{n}P_{r} = {}^{n}P_{r+1} \)
2. \( {}^{n}C_{r} = {}^{n}C_{r-1} \)

### Step 1: Analyze the first equation

The first equation states that the number of permutations of \( n \) items taken \( r \) at a time is equal to the number of permutations of \( n \) items taken \( r+1 \) at a time.

Using the formula for permutations:
\[ {}^{n}P_{r} = \frac{n!}{(n-r)!} \]
\[ {}^{n}P_{r+1} = \frac{n!}{(n-(r+1))!} = \frac{n!}{(n-r-1)!} \]

Setting these equal gives:
\[ \frac{n!}{(n-r)!} = \frac{n!}{(n-r-1)!} \]

### Step 2: Simplify the equation

We can cancel \( n! \) from both sides (assuming \( n \neq 0 \)):
\[ \frac{1}{(n-r)!} = \frac{1}{(n-r-1)!} \]

This implies:
\[ (n-r-1)! = (n-r)! \]

### Step 3: Solve for \( n \) and \( r \)

The above equation holds if \( n - r - 1 = 0 \), which leads to:
\[ n - r = 1 \quad \Rightarrow \quad n = r + 1 \tag{1} \]

### Step 4: Analyze the second equation

Now, we analyze the second equation:
\[ {}^{n}C_{r} = {}^{n}C_{r-1} \]

Using the formula for combinations:
\[ {}^{n}C_{r} = \frac{n!}{r!(n-r)!} \]
\[ {}^{n}C_{r-1} = \frac{n!}{(r-1)!(n-r+1)!} \]

Setting these equal gives:
\[ \frac{n!}{r!(n-r)!} = \frac{n!}{(r-1)!(n-r+1)!} \]

### Step 5: Simplify the second equation

Cancel \( n! \):
\[ \frac{1}{r!(n-r)!} = \frac{1}{(r-1)!(n-r+1)!} \]

This implies:
\[ (n-r+1)! = r \cdot (n-r)! \]

### Step 6: Substitute \( n \) from equation (1)

Substituting \( n = r + 1 \) into the equation:
\[ (n-r+1)! = r \cdot (n-r)! \]
\[ (1+1)! = r \cdot 1! \quad \Rightarrow \quad 2 = r \cdot 1 \]

Thus, we find:
\[ r = 2 \]

### Step 7: Find \( n \)

Using \( r = 2 \) in equation (1):
\[ n = r + 1 = 2 + 1 = 3 \]

### Conclusion

The values of \( n \) and \( r \) are:
\[ n = 3, \quad r = 2 \]

### Final Answer
The answer is \( n = 3 \) and \( r = 2 \).

---