Key Concepts and Formulas

• Permutation formula: ${}^n{P_r} = \frac{n!}{(n-r)!}$
• Combination formula: ${}^n{C_r} = \frac{n!}{r!(n-r)!}$
• Property of combinations: ${}^n{C_r} = {}^n{C_{n-r}}$

Step-by-Step Solution

Step 1: Use the permutation formula to simplify the first equation. We are given ${}^n{P_r} = {}^n{P_{r + 1}}$. Substituting the permutation formula, we have: $\frac{n!}{(n - r)!} = \frac{n!}{(n - (r + 1))!}$ Simplifying, $\frac{n!}{(n - r)!} = \frac{n!}{(n - r - 1)!}$ Since $n!$ is present on both sides, we can cancel it out (assuming $n! \neq 0$): $\frac{1}{(n - r)!} = \frac{1}{(n - r - 1)!}$ Taking the reciprocal of both sides, we get: $(n - r)! = (n - r - 1)!$ To make the factorials equal, we must have $n - r = 1$, since $x! = (x-1)!$ only if $x=1$. $n - r = 1$

Step 2: Use the combination formula to simplify the second equation. We are given ${}^n{C_r} = {}^n{C_{r - 1}}$. Substituting the combination formula, we have: $\frac{n!}{r!(n - r)!} = \frac{n!}{(r - 1)!(n - (r - 1))!}$ $\frac{n!}{r!(n - r)!} = \frac{n!}{(r - 1)!(n - r + 1)!}$ Since $n!$ is present on both sides, we can cancel it out (assuming $n! \neq 0$): $\frac{1}{r!(n - r)!} = \frac{1}{(r - 1)!(n - r + 1)!}$ Cross-multiplying, we have: $(r - 1)!(n - r + 1)! = r!(n - r)!$ $(r - 1)!(n - r + 1)(n - r)! = r(r - 1)!(n - r)!$ Now we can cancel out $(r-1)!$ and $(n-r)!$: $n - r + 1 = r$ $n + 1 = 2r$

Step 3: Solve the system of equations. From Step 1, we have $n - r = 1$, which can be rewritten as $n = r + 1$. From Step 2, we have $n + 1 = 2r$. Substitute $n = r + 1$ into the second equation: $(r + 1) + 1 = 2r$ $r + 2 = 2r$ $2 = r$ Therefore, $r = 2$.

Step 4: Verify the solution If $r = 2$, then $n = r + 1 = 2 + 1 = 3$. ${}^3{P_2} = \frac{3!}{(3-2)!} = \frac{6}{1} = 6$ ${}^3{P_3} = \frac{3!}{(3-3)!} = \frac{6}{1} = 6$ So, ${}^3{P_2} = {}^3{P_3}$ holds true.

${}^3{C_2} = \frac{3!}{2!(3-2)!} = \frac{6}{2 \cdot 1} = 3$ ${}^3{C_1} = \frac{3!}{1!(3-1)!} = \frac{6}{1 \cdot 2} = 3$ So, ${}^3{C_2} = {}^3{C_1}$ holds true.

However, the stated "Correct Answer" is 1, but our calculations show the correct answer is 2. Let's re-examine the solution for an error.

Step 1 (Revisited): We have $n-r = 1$. Step 2 (Revisited): We have $n+1 = 2r$. Substituting $n = r+1$ into $n+1 = 2r$ gives $(r+1) + 1 = 2r$, so $r+2 = 2r$, which means $r=2$.

If the correct answer is supposed to be 1, let's try to solve backwards. If $r=1$, then from $n-r=1$, we get $n=2$. Then, $n+1=2r$ implies $2+1 = 2(1)$, which gives $3=2$, a contradiction. So $r=1$ is impossible.

There must be an error in the problem statement or the "Correct Answer." Given the problem constraints, our solution yields $r=2$.

Let's proceed assuming the given answer key is incorrect and $r=2$ is the correct answer.

Common Mistakes & Tips

• Remember to carefully apply the definitions of permutations and combinations.
• When simplifying factorial expressions, look for common factors to cancel out.
• Always double-check your algebraic manipulations to avoid errors.

Summary

We used the permutation and combination formulas to set up a system of equations based on the given conditions. Solving this system, we found that $r = 2$. This value satisfies both given equations.

Final Answer The final answer is \boxed{2}, which corresponds to option (C).