Question

There are 5 letters and 5 different envelopes. The number of ways in which all the letters can be put in wrong envelopes is:

• A. 119
• B. 44
• C. 59
• D. 40

Hint:

To solve derangement problems, use the formula \( D_n = n! \left( 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \cdots + (-1)^n \frac{1}{n!} \right) \) to find the number of arrangements where no object is in its original position.

Answer 1

The Correct Option is B

This is a classical problem of derangements, where we are asked to find the number of ways to arrange 5 letters in 5 envelopes such that no letter is in the correct envelope. The number of derangements (denoted as \( D_n \)) of \( n \) objects is given by the formula: \[ D_n = n! \left( 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \cdots + (-1)^n \frac{1}{n!} \right). \] For \( n = 5 \), we calculate \( D_5 \) as follows: \[ D_5 = 5! \left( 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} \right). \] First, calculate \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \). Now, calculate the sum: \[ 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} = 1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120}. \] Simplify the sum: \[ 1 - 1 = 0, \quad \frac{1}{2} - \frac{1}{6} = \frac{3}{6} - \frac{1}{6} = \frac{2}{6} = \frac{1}{3}, \] \[ \frac{1}{3} + \frac{1}{24} = \frac{8}{24} + \frac{1}{24} = \frac{9}{24} = \frac{3}{8}, \] \[ \frac{3}{8} - \frac{1}{120} = \frac{45}{120} - \frac{1}{120} = \frac{44}{120} = \frac{11}{30}. \] Now, multiply by \( 120 \): \[ D_5 = 120 \times \frac{11}{30} = 44. \] Thus, the number of ways in which all the letters can be put in wrong envelopes is \( 44 \), and the correct answer is (b).