Question

There are 4 letters and 4 directed envelopes. The number of ways in which all the letters can be put in wrong envelope is

• A. 9
• B. 4
• C. 5
• D. 12

Answer 1

Answer: (A)

Let us first consider 2 letters and 2 envelopes, then there is only one way to place both the letters in wrong envelope.
Next, we consider 3 letters and 3 directed envelopes. The number of ways of putting all letters in wrong envelopes
= Total number of possible arrangements - Number of ways in which all letters are in correct envelopes
- Number of ways in which 1 letter in correct envelope $=3!-1{-}^3{C}_1\times 1=2$
[ $\therefore$ The case of two letters in correct envelope and one in wrong envelope is not possible]
Further, we consider 4 letters and 4 directed envelopes. The number of ways of putting all letters in wrong envelopes
= Total number of possible arrangements - number of ways in which all letters are in correct envelope - Number of ways in which 1 letter is in correct envelopes (3 in wrong envelope)
- Number of ways in which 2 letters are in correct envelope (2 in wrong envelope) $=4!-1{-}^4{C}_1\times 1=9$
NOTE : Such problems are called problems of deragement. Hence, using the formula of deragement. The required number of ways of placing all letters in wrong envelope
$4!\left[1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{2!}+\frac{1}{4!}\right]=\frac{4!}{2!}-\frac{4!}{3!}+\frac{4!}{4!}=12-4+1=9$