Key Concepts and Formulas

This problem relies on the fundamental concepts of the Binomial Theorem, specifically the formula for the general term in an expansion and understanding terms from the end of an expansion.

1. General Term in Binomial Expansion: For an expansion $(a+b)^n$, the $(r+1)^{\text{th}}$ term from the beginning, denoted as $T_{r+1}$, is given by: $T_{r+1} = {}^n C_r a^{n-r} b^r$ where ${}^n C_r = \frac{n!}{r!(n-r)!}$ is the binomial coefficient.

2. $k^{\text{th}}$ Term from the End: The $k^{\text{th}}$ term from the end in the expansion of $(a+b)^n$ is equivalent to the $(n-k+2)^{\text{th}}$ term from the beginning. Alternatively, due to the symmetry of binomial coefficients (${}^n C_r = {}^n C_{n-r}$), the $k^{\text{th}}$ term from the end of $(a+b)^n$ can be found by treating it as the $k^{\text{th}}$ term from the beginning of $(b+a)^n$, i.e., $T'_{k} = {}^n C_{k-1} b^{n-(k-1)} a^{k-1}$.

Step-by-step Derivation

Let the given expansion be $\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^n$.

1. Identify $a$ and $b$ and the General Term Formula

From the given expression, we identify the terms $a$ and $b$: $a = \sqrt[3]{2} = 2^{1/3}$ $b = \frac{1}{\sqrt[3]{3}} = 3^{-1/3}$

2. Determine the $15^{\text{th}}$ Term from the Beginning ($T_{15}$)

To find the $15^{\text{th}}$ term, we set $r+1 = 15$, which means $r = 14$. Using the general term formula $T_{r+1} = {}^n C_r a^{n-r} b^r$: $T_{15} = {}^n C_{14} (2^{1/3})^{n-14} (3^{-1/3})^{14}$ $T_{15} = {}^n C_{14} 2^{(n-14)/3} 3^{-14/3}$

3. Determine the $15^{\text{th}}$ Term from the End ($T'_{15}$)

The $15^{\text{th}}$ term from the end in the expansion of $(a+b)^n$ is the $(n-15+2)^{\text{th}} = (n-13)^{\text{th}}$ term from the beginning. For the $(n-13)^{\text{th}}$ term from the beginning, we set $r+1 = n-13$, so $r = n-14$. Using the general term formula: $T'_{15} = {}^n C_{n-14} a^{n-(n-14)} b^{n-14}$ $T'_{15} = {}^n C_{n-14} a^{14} b^{n-14}$ Since ${}^n C_{n-r} = {}^n C_r$, we know that ${}^n C_{n-14} = {}^n C_{14}$. Substituting this and the values of $a$ and $b$: $T'_{15} = {}^n C_{14} (2^{1/3})^{14} (3^{-1/3})^{n-14}$ $T'_{15} = {}^n C_{14} 2^{14/3} 3^{-(n-14)/3}$

4. Set up and Simplify the Ratio of the Terms

We are given that the ratio of the $15^{\text{th}}$ term from the beginning to the $15^{\text{th}}$ term from the end is $\frac{1}{6}$. $\frac{T_{15}}{T'_{15}} = \frac{1}{6}$ Substitute the expressions for $T_{15}$ and $T'_{15}$: $\frac{{}^n C_{14} 2^{(n-14)/3} 3^{-14/3}}{{}^n C_{14} 2^{14/3} 3^{-(n-14)/3}} = \frac{1}{6}$ The binomial coefficients ${}^n C_{14}$ cancel out. Group terms with the same base: $2^{\frac{n-14}{3} - \frac{14}{3}} \cdot 3^{-\frac{14}{3} - (-\frac{n-14}{3})} = \frac{1}{6}$ $2^{\frac{n-14-14}{3}} \cdot 3^{\frac{-14 + (n-14)}{3}} = \frac{1}{6}$ $2^{\frac{n-28}{3}} \cdot 3^{\frac{n-28}{3}} = \frac{1}{6}$ Combine the terms with the same exponent: $(2 \cdot 3)^{\frac{n-28}{3}} = \frac{1}{6}$ $6^{\frac{n-28}{3}} = 6^{-1}$

5. Solve for $n$

Since the bases are equal, we can equate the exponents: $\frac{n-28}{3} = -1$ Multiply both sides by 3: $n-28 = -3$ Add 28 to both sides: $n = 28 - 3$ $n = 25$

6. Calculate the Value of ${}^n \mathrm{C}_3$

Now that we have the value of $n=25$, we need to calculate ${}^{25} C_3$. ${}^{25} C_3 = \frac{25!}{3!(25-3)!} = \frac{25!}{3!22!}$ ${}^{25} C_3 = \frac{25 \times 24 \times 23}{3 \times 2 \times 1}$ ${}^{25} C_3 = 25 \times 4 \times 23$ ${}^{25} C_3 = 100 \times 23$ ${}^{25} C_3 = 2300$

Important Tips and Common Mistakes

• Understanding Terms from the End: A common mistake is to assume the $k^{\text{th}}$ term from the end simply swaps $a$ and $b$ in the $k^{\text{th}}$ term from the beginning. While this often works for the ratio due to symmetry, it's safer to convert the term from the end to its equivalent term from the beginning (i.e., the $k^{\text{th}}$ term from the end is the $(n-k+2)^{\text{th}}$ term from the beginning).
• Careful with Exponents: Pay close attention to the rules of exponents when simplifying ratios, especially with fractional and negative exponents.
• Simplification of $\frac{a}{b}$: Simplify the base expression $\left(\frac{a}{b}\right)$ as much as possible before raising it to the power. Here, $\frac{2^{1/3}}{3^{-1/3}} = 2^{1/3} \cdot 3^{1/3} = (2 \cdot 3)^{1/3} = 6^{1/3}$.
• Checking Options: If your calculated $n$ leads to an answer not among the options, re-check your calculations. In this case, our derived answer (2300) corresponds to option (B). The stated "Correct Answer: A" (4960) would imply $n=32$, which is not derived from the problem statement as written. It is important to trust your derivation based on the problem's explicit conditions.

Summary and Key Takeaway

By correctly applying the general term formula for binomial expansion and understanding how to determine terms from the end, we systematically simplified the given ratio. Equating the simplified ratio to the given value allowed us to solve for $n$. Finally, using the calculated $n$, we computed the binomial coefficient ${}^n C_3$. The key takeaway is the meticulous application of exponent rules and binomial term definitions to avoid algebraic errors. The calculation yields $n=25$, resulting in ${}^{25} C_3 = 2300$.