Key Concepts and Formulas

• Derangement: A derangement of a set is a permutation in which no element appears in its original position.
• Formula for Derangements: The number of derangements of $n$ objects, denoted by $D_n$ or $!n$, is given by: $D_n = n! \sum_{k=0}^{n} \frac{(-1)^k}{k!} = n! \left( \frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \cdots + \frac{(-1)^n}{n!} \right)$
• For small values of $n$, derangements can also be calculated recursively or using pre-calculated values. $D_0 = 1, D_1 = 0, D_2 = 1, D_3 = 2, D_4 = 9, D_5 = 44$.

Step-by-Step Solution

• Step 1: Identify the problem as a derangement problem. We are given that 5 students have been allotted seats according to their roll numbers. We need to find the number of ways in which none of the students sits in their allotted seat. This is a classic derangement problem where we want to find the number of permutations where no student occupies their original seat.

• Step 2: Apply the formula for derangements with $n=5$. We need to find $D_5$, the number of derangements of 5 objects. Using the formula: $D_5 = 5! \left( \frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} \right)$

• Step 3: Calculate the factorials and simplify the expression. $D_5 = 120 \left( 1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120} \right)$ $D_5 = 120 \left( \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120} \right)$

• Step 4: Find a common denominator and combine the fractions. The least common denominator is 120. $D_5 = 120 \left( \frac{60}{120} - \frac{20}{120} + \frac{5}{120} - \frac{1}{120} \right)$ $D_5 = 120 \left( \frac{60 - 20 + 5 - 1}{120} \right)$ $D_5 = 120 \left( \frac{44}{120} \right)$

• Step 5: Simplify to find the final answer. $D_5 = 44$

Common Mistakes & Tips

• Misidentifying the problem: Make sure to recognize the problem as a derangement before applying the formula. Look for keywords like "none in their original position".
• Incorrectly applying the formula: Be careful with the signs in the derangement formula. Ensure that the terms alternate between positive and negative.
• Calculation errors: Double-check your arithmetic, especially when dealing with factorials and fractions.

Summary

The problem asks for the number of ways in which none of the 5 students sit in their allotted seat, which is a derangement problem. We used the formula for derangements, $D_n = n! \sum_{k=0}^{n} \frac{(-1)^k}{k!}$, and substituted $n=5$. After calculating the factorials and simplifying the expression, we found that $D_5 = 44$. Therefore, there are 44 ways in which none of the students sit on the allotted seat.

Final Answer

The final answer is \boxed{44}.