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Question

Question: A person writes n letters and addresses on envelopes; if the letters are placed in the envelopes at ...

A person writes n letters and addresses on envelopes; if the letters are placed in the envelopes at random, what is the probability that every letter goes wrong?

Explanation

Solution

Hint: Here we will proceed by assuming un{u_n}as the required answer i.e. number of ways in which all the letters go wrong. Then we will find the required probability by formula - un=(n1)(un1+un2){u_n} = \left( {n - 1} \right)\left( {{u_{n - 1}} + {u_{n - 2}}} \right).

Complete step-by-step answer:

Let un{u_n} deNote:the number of ways in which all the letters go wrong
Let a, b, c, d…. represent that arrangement in which all the letters are in their own envelopes.
Now if aa in any other arrangement occupies the place of an assigned letter bb, this letter must either occupy a’s place or some other.

1. Suppose bb occupies a’s place. Then the number of ways in which all the remaining n-2 letters can be displaced is un2{u_{n - 2}}.
2. Therefore the numbers of ways in which aamay be displaced by interchange with some one of the other n-1 letters, and the rest be all displaced is (n1)un1\left( {n - 1} \right){u_{n - 1}}.
3. Suppose aa occupies b’s place, and bb does not occupy a’s place. Then in arrangements satisfying the required conditions, since aais fixed in b’s place, the letter b, c, d, ……. must be all displaced, which can be done in un1{u_{n - 1}} ways.
4. Therefore the number of ways in which aaoccupies the place of another letter but not by interchange with that letter is-
\therefore un=(n1)(un1+un2){u_n} = \left( {n - 1} \right)\left( {{u_{n - 1}} + {u_{n - 2}}} \right) ;
Now we will find unnun1=(1)n(u2u1){u_n} - n{u_{n - 1}} = {\left( { - 1} \right)^n}\left( {{u_2} - {u_1}} \right)
Also u1=0,u2=1{u_1} = 0,{u_2} = 1
Thus, we finally obtain-
{u_n} = n\left\\{ {\dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{4} - ..... + \dfrac{{{{\left( { - 1} \right)}^n}}}{n}} \right\\}
Now the total number of ways in which the nn things can be put in nn places,
Therefore, the required chance is –
1213+14.....+(1)nn\dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{4} - ..... + \dfrac{{{{\left( { - 1} \right)}^n}}}{n}

Note: Here in this question, we can assume any variable instead of x to find the probability of every letter goes wrong. As a and b are our assumed variables for converting given conditions, we can use any other variables so that we can find the required answer using the given conditions.