If `^n P_r=^n P_(r+1)a n d^n C_r=^n C_(r-1,)` then the value of `n+r` is.

Options

A. 2
B. 3
C. 4
D. 5

Answer

To solve the problem, we need to use the given equations involving permutations and combinations:

1. **Given equations:**
\[ ^n P_r = ^n P_{r+1} \]
\[ ^n C_r = ^n C_{r-1} \]

2. **Step 1: Solve the first equation \( ^n P_r = ^n P_{r+1} \)**

The formula for permutations is given by:
\[ ^n P_r = \frac{n!}{(n-r)!} \]
and
\[ ^n P_{r+1} = \frac{n!}{(n-(r+1))!} = \frac{n!}{(n-r-1)!} \]

Setting the two expressions equal:
\[ \frac{n!}{(n-r)!} = \frac{n!}{(n-r-1)!} \]

Canceling \( n! \) from both sides:
\[ \frac{1}{(n-r)!} = \frac{1}{(n-r-1)!} \]

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

So, we can write:
\[ n = r + 1 \quad \text{(Equation 1)} \]

3. **Step 2: Solve the second equation \( ^n C_r = ^n C_{r-1} \)**

The formula for combinations is given by:
\[ ^n C_r = \frac{n!}{r!(n-r)!} \]
and
\[ ^n C_{r-1} = \frac{n!}{(r-1)!(n-(r-1))!} = \frac{n!}{(r-1)!(n-r+1)!} \]

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

Canceling \( n! \) from both sides:
\[ \frac{1}{r!(n-r)!} = \frac{1}{(r-1)!(n-r+1)!} \]

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

Simplifying gives:
\[ (n-r+1) = r \implies n - 2r + 1 = 0 \]

Rearranging, we find:
\[ n = 2r - 1 \quad \text{(Equation 2)} \]

4. **Step 3: Solve the two equations simultaneously**

From Equation 1:
\[ n = r + 1 \]

Substitute \( n \) from Equation 1 into Equation 2:
\[ r + 1 = 2r - 1 \]

Rearranging gives:
\[ 1 + 1 = 2r - r \implies 2 = r \]

Now substituting \( r = 2 \) back into Equation 1:
\[ n = 2 + 1 = 3 \]

5. **Step 4: Find \( n + r \)**

Now we can find \( n + r \):
\[ n + r = 3 + 2 = 5 \]

Thus, the final answer is:
\[ \boxed{5} \]