Key Concepts and Formulas

• Pascal's Identity: ${}^n{C_r} + {}^n{C_{r - 1}} = {}^{n + 1}{C_r}$
• Combination Formula: ${}^n{C_r} = \frac{n!}{r!(n-r)!}$

Step-by-Step Solution

1. Strategic Rearrangement of the Expression

   The given expression is: ${}^n{C_{r + 1}} + {}^n{C_{r - 1}} + 2\, \times \,{}^n{C_r}$ We aim to use Pascal's Identity. To do that, we need to split the term $2\, \times \,{}^n{C_r}$ into two identical terms: ${}^n{C_r} + {}^n{C_r}$. This allows us to group terms in a way that Pascal's Identity can be applied. Rewriting the expression: ${}^n{C_{r + 1}} + {}^n{C_r} + {}^n{C_r} + {}^n{C_{r - 1}}$ Now, we can regroup the terms as follows: $= \left( {}^n{C_{r + 1}} + {}^n{C_r} \right) + \left( {}^n{C_r} + {}^n{C_{r - 1}} \right)$

2. First Application of Pascal's Identity

   We apply Pascal's Identity to each of the groups created in the previous step.

   • Group 1: Consider ${}^n{C_{r + 1}} + {}^n{C_r}$. We can rewrite this as ${}^n{C_{r}} + {}^n{C_{r+1}}$. Comparing this to the identity ${}^N{C_K} + {}^N{C_{K - 1}} = {}^{N + 1}{C_K}$, we can identify $N=n$ and $K=r+1$. The second term $C_{r+1}$ corresponds to $C_{K}$. The term $C_r$ corresponds to $C_{K-1}$. Therefore, applying Pascal's Identity to Group 1 gives: ${}^n{C_{r}} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}$

   • Group 2: Consider ${}^n{C_r} + {}^n{C_{r - 1}}$. This directly matches the identity ${}^N{C_K} + {}^N{C_{K - 1}} = {}^{N + 1}{C_K}$ with $N=n$ and $K=r$. Therefore, applying Pascal's Identity to Group 2 gives: ${}^n{C_r} + {}^n{C_{r - 1}} = {}^{n + 1}{C_r}$

   Now, substitute these simplified forms back into our expression: $= \left( {}^n{C_{r + 1}} + {}^n{C_r} \right) + \left( {}^n{C_r} + {}^n{C_{r - 1}} \right)$ $= {}^{n + 1}{C_{r + 1}} + {}^{n + 1}{C_r}$

3. Second Application of Pascal's Identity

   We are left with the expression: ${}^{n + 1}{C_{r + 1}} + {}^{n + 1}{C_r}$ Again, this expression perfectly fits the form of Pascal's Identity! We can rewrite it as: ${}^{n + 1}{C_{r}} + {}^{n + 1}{C_{r + 1}}$ Here, the 'n' value in the identity becomes $n+1$, and the 'r' value corresponds to $r+1$. So, comparing ${}^{N}{C_K} + {}^{N}{C_{K - 1}} = {}^{N + 1}{C_K}$ with ${}^{n + 1}{C_{r}} + {}^{n + 1}{C_{r + 1}}$, we have $N = n+1$ and $K = r+1$.

   Applying Pascal's Identity for the second time: ${}^{n + 1}{C_{r}} + {}^{n + 1}{C_{r + 1}} = {}^{(n + 1) + 1}{C_{r + 1}}$ $= {}^{n + 2}{C_{r + 1}}$

This matches option (C).

Common Mistakes & Tips

• Incorrect Application of Pascal's Identity: Ensure you correctly identify the 'n' and 'r' values when applying Pascal's Identity. Double-check that the 'n' values are the same and the 'r' values differ by exactly 1.
• Forgetting to Split the Term: The $2\, \times \,{}^n{C_r}$ term is a key indicator to split it and apply Pascal's Identity twice. Don't overlook this crucial step.

Summary

By strategically splitting the $2 \times {}^n{C_r}$ term and applying Pascal's Identity twice, we simplified the given expression ${}^n{C_{r + 1}} + {}^n{C_{r - 1}} + 2\, \times \,{}^n{C_r}$ to ${}^{n + 2}{C_{r + 1}}$.

Final Answer

The final answer is \boxed{{}^{n + 2}{C_{r + 1}}}, which corresponds to option (C).