If $$^nP_r = ^nP_{r+1}$$ and $$^nC_r = ^nC_{r-1}$$, then the value of $$r$$ is equal to:

We have the two relations

$$^nP_r \;=\; ^nP_{\,r+1}\qquad\text{and}\qquad ^nC_r \;=\; ^nC_{\,r-1}.$$

First we use the permutation formula

$$^nP_k \;=\; \frac{n!}{(n-k)!}.$$

Substituting $$k=r$$ and $$k=r+1$$ we obtain

$$^nP_r \;=\; \frac{n!}{(n-r)!},\qquad ^nP_{\,r+1} \;=\; \frac{n!}{(n-r-1)!}.$$

Since these two are equal, their quotient is $$1$$:

$$\frac{^nP_r}{^nP_{\,r+1}} \;=\; \frac{\dfrac{n!}{(n-r)!}}{\dfrac{n!}{(n-r-1)!}} \;=\; \frac{(n-r-1)!}{(n-r)!} \;=\; \frac{1}{n-r} \;=\; 1.$$

Thus

$$n-r \;=\; 1\;\;\Longrightarrow\;\; r \;=\; n-1. \quad -(1)$$

Next we use the combination formula

$$^nC_k \;=\; \frac{n!}{k!\,(n-k)!}.$$

Putting $$k=r$$ and $$k=r-1$$ we get

$$^nC_r \;=\; \frac{n!}{r!\,(n-r)!},\qquad ^nC_{\,r-1} \;=\; \frac{n!}{(r-1)!\,(n-r+1)!}.$$

Because these two are equal, their quotient is also $$1$$:

$$\frac{^nC_r}{^nC_{\,r-1}} \;=\; \frac{\dfrac{n!}{r!\,(n-r)!}} {\dfrac{n!}{(r-1)!\,(n-r+1)!}} \;=\; \frac{(r-1)!\,(n-r+1)!}{r!\,(n-r)!} \;=\; \frac{n-r+1}{r} \;=\; 1.$$

Hence

$$n-r+1 \;=\; r \;\;\Longrightarrow\;\; n+1 \;=\; 2r \;\;\Longrightarrow\;\; r \;=\; \frac{n+1}{2}. \quad -(2)$$

Now both (1) and (2) must hold simultaneously. Equating the two expressions for $$r$$ we write

$$n-1 \;=\; \frac{n+1}{2}.$$

Multiplying by $$2$$ and simplifying, we have

$$2n-2 \;=\; n+1 \;\;\Longrightarrow\;\; n-3 \;=\; 0 \;\;\Longrightarrow\;\; n \;=\; 3.$$

Finally, substituting $$n=3$$ into $$r=n-1$$ gives

$$r \;=\; 3-1 \;=\; 2.$$

Therefore

$$r = 2.$$

Hence, the correct answer is Option C.