There are 5 letters and 5 addressed envelopes. The number of ways in which the letters can be placed in the envelopes so that none of them goes into the right envelope is
A. 22
B. 44
C. 120
D. 119
Answer
558.6k+ views
Hint: In this question we have to make sure no letter goes into the right envelope. For this, find the number of ways in which all goes in the correct envelope, only 1 letter goes in the correct envelope, 2 letters go in the correct envelope, 3 letters go in the correct envelope, 4 letters go in the correct envelope and subtract all of them from the total number of ways without restrictions.
Complete step-by-step answer:
There are 5 letters and 5 addressed envelopes.
The number of ways of posting letters = 5!
Number of ways in which all letters are in correct envelope = 1
Number of ways in which only 1 letter goes into correct envelope i.e., 1 letter goes correctly in \[{}^5{C_1}\] ways and the other 4 letters goes incorrectly in 9 ways = \[{}^5{C_1} \times 9\]
Number of ways in which only 2 letter goes into correct envelope i.e., 2 letters goes correctly in \[{}^5{C_2}\] ways and the other 3 letters goes incorrectly in 2 ways \[ = {}^5{C_2} \times 2\]
Number of ways in which only 3 letter goes into correct envelope i.e., 3 letters goes correctly in \[{}^5{C_3}\] ways and the other 2 letters goes incorrectly in 1 way \[ = {}^5{C_3} \times 1\]
Number of ways in which only 4 letters goes into the correct envelope = 0. Because if 4 letters go correctly, then the remaining one will also go into the correct envelope.
Total number of ways to put 3 letters and 5 envelopes so that all goes in wrong envelope
\[
= 5! - 1 - {}^5{C_1} \times 9 - {}^5{C_2} \times - {}^5{C_3} \times 1 - 0 \\
= 120 - 1 - 45 - 20 - 10 - 0 \\
= 44 \\
\]
So, the correct answer is “Option B”.
Note: This problem can also be solved by a basic method of counting. But the use of permutation and combination in these types of problems makes the task easier. Permutation gives the total no of arrangements possible and combination gives total number of selection possible.
Complete step-by-step answer:
There are 5 letters and 5 addressed envelopes.
The number of ways of posting letters = 5!
Number of ways in which all letters are in correct envelope = 1
Number of ways in which only 1 letter goes into correct envelope i.e., 1 letter goes correctly in \[{}^5{C_1}\] ways and the other 4 letters goes incorrectly in 9 ways = \[{}^5{C_1} \times 9\]
Number of ways in which only 2 letter goes into correct envelope i.e., 2 letters goes correctly in \[{}^5{C_2}\] ways and the other 3 letters goes incorrectly in 2 ways \[ = {}^5{C_2} \times 2\]
Number of ways in which only 3 letter goes into correct envelope i.e., 3 letters goes correctly in \[{}^5{C_3}\] ways and the other 2 letters goes incorrectly in 1 way \[ = {}^5{C_3} \times 1\]
Number of ways in which only 4 letters goes into the correct envelope = 0. Because if 4 letters go correctly, then the remaining one will also go into the correct envelope.
Total number of ways to put 3 letters and 5 envelopes so that all goes in wrong envelope
\[
= 5! - 1 - {}^5{C_1} \times 9 - {}^5{C_2} \times - {}^5{C_3} \times 1 - 0 \\
= 120 - 1 - 45 - 20 - 10 - 0 \\
= 44 \\
\]
So, the correct answer is “Option B”.
Note: This problem can also be solved by a basic method of counting. But the use of permutation and combination in these types of problems makes the task easier. Permutation gives the total no of arrangements possible and combination gives total number of selection possible.
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