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Q.
There are 4 letters and 4 directed envelopes. The number of ways in which all the letters can be put in wrong envelope is
Permutations and Combinations
Solution:
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 $= 3 \,! - 1 - ^3C_1 × 1 = 2$
[ $\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) $= 4\, ! - 1 - ^4C_1 × 1 = 9$ 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
$4!\left[1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{2!}+\frac{1}{4!}\right]=\frac{4!}{2!}-\frac{4!}{3!}+\frac{4!}{4!}=12-4+1=9$