\(\frac{^nC_4 (2^\frac{1}{4})^{n-4}(3^{\frac{-1}{4}})^4}{^nC_4 (2^\frac{-1}{4})^{n-4}(3^{\frac{1}{4}})^4}\) = \(\sqrt 6\)
\((\frac{2^\frac{1}{4}}{3^\frac{-1}{4}})^{(n-8)}\) = \(\sqrt 6\)
\((6)^\frac{n-8}{4}\) = \(\sqrt 6\)
\(n-8=2\)
\(n=10\)
\(T_3 = {^{10}}C_2(2^{\frac{1}{4}})^8 (3^\frac{-1}{4})^2\)
\(= {^{10}}C_2\times(\sqrt2)^4\times\frac{1}{\sqrt3} = 60\sqrt3\)
So, the correct answer is (B): \(60\sqrt 3\)
\[ \left( \frac{1}{{}^{15}C_0} + \frac{1}{{}^{15}C_1} \right) \left( \frac{1}{{}^{15}C_1} + \frac{1}{{}^{15}C_2} \right) \cdots \left( \frac{1}{{}^{15}C_{12}} + \frac{1}{{}^{15}C_{13}} \right) = \frac{\alpha^{13}}{{}^{14}C_0 \, {}^{14}C_1 \cdots {}^{14}C_{12}} \]
Then \[ 30\alpha = \underline{\hspace{1cm}} \]
Method used for separation of mixture of products (B and C) obtained in the following reaction is: 
In the following \(p\text{–}V\) diagram, the equation of state along the curved path is given by \[ (V-2)^2 = 4ap, \] where \(a\) is a constant. The total work done in the closed path is: 
Let \( ABC \) be a triangle. Consider four points \( p_1, p_2, p_3, p_4 \) on the side \( AB \), five points \( p_5, p_6, p_7, p_8, p_9 \) on the side \( BC \), and four points \( p_{10}, p_{11}, p_{12}, p_{13} \) on the side \( AC \). None of these points is a vertex of the triangle \( ABC \). Then the total number of pentagons that can be formed by taking all the vertices from the points \( p_1, p_2, \ldots, p_{13} \) is ___________.
The binomial theorem formula is used in the expansion of any power of a binomial in the form of a series. The binomial theorem formula is
