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.
To determine the 315th word in the dictionary arrangement of the letters of the word "NAGPUR", we follow these steps:
We calculate the total number of permutations starting with each letter until we reach the desired position:
Now, calculate from NP to find position 315:
Continuing with NRA (R can be followed by A, G, P, U forming 4! each):
Thus, the word at the 315th position is NRAPGU.
The number of strictly increasing functions \(f\) from the set \(\{1, 2, 3, 4, 5, 6\}\) to the set \(\{1, 2, 3, ...., 9\}\) such that \(f(i)>i\) for \(1 \le i \le 6\), is equal to:
Method used for separation of mixture of products (B and C) obtained in the following reaction is: 