\[ \frac{{n+1}C_r}{nC_r} = \frac{55}{35} \]
\[ \frac{(n+1)!}{(r+1)!(n-r)!} \div \frac{n!}{r!(n-r)!} = \frac{11}{7} \]
\[ \frac{n+1}{r+1} = \frac{11}{7} \]
\[ 7n = 4 + 11r \]
\[ \frac{nC_r}{n-1C_{r-1}} = \frac{35}{21} \]
\[ \frac{n!}{r!(n-r)!} \div \frac{(n-1)!}{(r-1)!(n-r)!} = \frac{5}{3} \]
\[ \frac{n}{r} = \frac{5}{3} \]
\[ 3n = 5r \]
Solving for \( r = 6 \) and \( n = 10 \):
\[ 2n + 5r = 50 \]
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32