
There are four letters and four directed envelopes. The number of ways in which all the letters can be put in the wrong envelope is
(a) 8
(b) 9
(c) 16
(d) None of these
Answer
507.1k+ views
Hint: At first, find all the possible ways and then subtract it with ways which do not satisfy the conditions like 1 letter goes to correct envelopes, 2 letters goes to the correct envelope and 4 letters goes to the correct envelope as 3 letters is not possible.
Complete step-by-step answer:
In the question, we are given four letters and four direct envelopes, we have to find the number of ways so that all the letters are put in the wrong envelope.
Now by the given wording in the question if any one letter goes to the right envelope then it counts as fails.
So, we will count first all the ways of failure and then subtract it from all the cases of inserting letters in envelopes which is 4! or 24 ways.
The failure will occur if exactly 1 letter is in the correct envelope. Since there are 4 envelopes so there are 4 different values to accomplish this. Each of the other letters must be in the wrong envelope and each of the three has 1 choice of 2 possible wrong letters. So, the total ways are $4\times 2$ or 8 for this situation.
The failure will also occur if two letters are put in two wrong envelopes and two in two right envelopes, which can be done in ${}^{4}{{C}_{2}}$ or 6 ways.
For three letters to correct envelopes is not possible because the fourth one would necessarily go in the correct one.
Now, there are only ${}^{4}{{C}_{4}}$ or only 1 way to correct all envelopes.
So the total number of ways of failure is 8+6+1 or 15 ways.
So, the ways to satisfy the given condition of the Question is $24-15$ or 9.
Hence, the correct option is ‘b’.
Note: There is also a shortcut or a formula to solve this problem which is $g\left( n \right)=n!-\left( 1+\sum\limits_{k=1}^{n-2}{\left[ {}^{n}{{C}_{k}}\times g\left( n-k \right) \right]} \right)$ where n is total number of envelopes $g\left( n \right)$ be total number of ways and k is values from 1 to $\left( n-2 \right)$ .
Complete step-by-step answer:
In the question, we are given four letters and four direct envelopes, we have to find the number of ways so that all the letters are put in the wrong envelope.
Now by the given wording in the question if any one letter goes to the right envelope then it counts as fails.
So, we will count first all the ways of failure and then subtract it from all the cases of inserting letters in envelopes which is 4! or 24 ways.
The failure will occur if exactly 1 letter is in the correct envelope. Since there are 4 envelopes so there are 4 different values to accomplish this. Each of the other letters must be in the wrong envelope and each of the three has 1 choice of 2 possible wrong letters. So, the total ways are $4\times 2$ or 8 for this situation.
The failure will also occur if two letters are put in two wrong envelopes and two in two right envelopes, which can be done in ${}^{4}{{C}_{2}}$ or 6 ways.
For three letters to correct envelopes is not possible because the fourth one would necessarily go in the correct one.
Now, there are only ${}^{4}{{C}_{4}}$ or only 1 way to correct all envelopes.
So the total number of ways of failure is 8+6+1 or 15 ways.
So, the ways to satisfy the given condition of the Question is $24-15$ or 9.
Hence, the correct option is ‘b’.
Note: There is also a shortcut or a formula to solve this problem which is $g\left( n \right)=n!-\left( 1+\sum\limits_{k=1}^{n-2}{\left[ {}^{n}{{C}_{k}}\times g\left( n-k \right) \right]} \right)$ where n is total number of envelopes $g\left( n \right)$ be total number of ways and k is values from 1 to $\left( n-2 \right)$ .
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