Arranging the letters in alphabetical order: NAGPUR
Starting with \( A \): \( 5! = 120 \) positions
Starting with \( G \): \( 5! = 120 \) positions, cumulative: 240
Starting with \( N \) and \( A \): \( 4! = 24 \) positions, cumulative: 264
Starting with \( N \) and \( G \): \( 4! = 24 \) positions, cumulative: 288
Starting with \( N \) and \( P \): \( 4! = 24 \) positions, cumulative: 312
Now, starting with \( N \), \( R \), and \( A \):
\[ \text{NRAGUP} = 1, \text{ cumulative: 313} \]
\[ \text{NRAGPU} = 1, \text{ cumulative: 314} \]
\[ \text{NRAPGU} = 1, \text{ cumulative: 315} \]
Thus, the word at the \( 315^{\text{th}} \) position is NRAPGU.
In the circuit shown, assuming the threshold voltage of the diode is negligibly small, then the voltage \( V_{AB} \) is correctly represented by:
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: