In the expansion of $\bigl(\frac{\sqrt{2}}{3} + \frac{1}{\sqrt{3}}\bigr)^n$, n $\in \mathbb{N}$ if the ratio of the 15th term from the beginning to the 15th term from the end is $\tfrac{1}{6}$ then the value of $\binom{n}{3}$ is:

Step 1: Identify the terms in the binomial expansion

The general term in the binomial expansion of $(a+b)^n$ is given by the formula $T_{r+1} = \binom{n}{r} a^{n-r} b^r$. In this problem, $a = \frac{\sqrt{2}}{3}$ and $b = \frac{1}{\sqrt{3}}$.

Step 2: Express the 15th term from the beginning and the end

The 15th term from the beginning is $T_{15}$, which corresponds to $r=14$. The 15th term from the end is $T'_{15}$. In a binomial expansion of $n+1$ terms, the $k$ -th term from the end is the $(n-k+2)$ -th term from the beginning. So, the 15th term from the end is the $(n-15+2)$ -th term from the beginning, which is $T_{n-13}$. Using the property $\binom{n}{k} = \binom{n}{n-k}$, we can write $T_{n-13}$ as $\binom{n}{n-14} a^{14} b^{n-14}$.

Step 3: Set up the ratio and simplify

We are given that the ratio of the 15th term from the beginning to the 15th term from the end is $\frac{1}{6}$. We substitute the expressions for $T_{15}$ and $T'_{15}$ into this ratio. Since $\binom{n}{14} = \binom{n}{n-14}$, these terms cancel out.

Step 4: Solve for n

After canceling the binomial coefficients, we combine the terms with the same base. This simplifies to $\left(\frac{\sqrt{2}/3}{1/\sqrt{3}}\right)^{n-28} = \frac{1}{6}$. Further simplification leads to $\left(\frac{\sqrt{6}}{3}\right)^{n-28} = \frac{1}{6}$. Recognizing that $\frac{1}{6} = \left(\frac{\sqrt{6}}{3}\right)^{-2}$, we equate the exponents to find $n-28 = -2$, which gives $n=26$.

Step 5: Calculate the value of $\binom{n}{3}$

Now that we have found $n=26$, we need to calculate the value of $\binom{n}{3}$. Substituting $n=26$, we get $\binom{26}{3} = \frac{26!}{3!(26-3)!} = \frac{26 \times 25 \times 24}{3 \times 2 \times 1}$. Performing the multiplication and division gives the final result.

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