Let the 8 boxes be arranged in three rows as shown:
Let \( R_1, R_2, R_3 \) represent the three rows. \( R_1 \to \) (1st row), \( R_2 \to \) (2nd row), \( R_3 \to \) (3rd row). Total number of ways: \[ \text{Total} = \left[ (\text{All in } R_1 \text{ and } R_3) + (\text{All in } R_2 \text{ and } R_3) + (\text{All in } R_1 \text{ and } R_2) \right] \] \[ = 8C5 \times 5! - \left[ \text{(ways to place in 1st and 2nd row)} + \text{(ways to place in 3rd row)} \right] \]
\[ = \left| (56-1) \times 6 \right| = 120 \times 48 = 5760 \] Hence, the total number of ways to arrange the letters is \( 5760 \).
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).