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Question: A person writes \(4\) letters and addresses \(4\) envelopes. If the letters are placed in the envelo...

A person writes 44 letters and addresses 44 envelopes. If the letters are placed in the envelopes at random, what is the probability that all letters are not placed in the right envelopes?
A. 14\dfrac{1}{4}
B. 1124\dfrac{{11}}{{24}}
C. 1524\dfrac{{15}}{{24}}
D. 2324\dfrac{{23}}{{24}}

Explanation

Solution

The number of ways in which 44 envelopes can be placed in the 44 envelopes =4!=24 = 4! = 24
And if all the letters go into the right envelopes so that can be done only in 11 way
So the probability of getting all the letters in the right envelopes =124 = \dfrac{1}{{24}}
Hence we can also find the probability of not placing all the letters in the right envelopes.

Complete step by step solution:
Here we are given in the question that person writes 44 letters and addresses 44 envelopes. Let us assume that the four letters he wrote be represented by L1,L2,L3,L4{L_1},{L_2},{L_3},{L_4} and the envelopes be E1,E2,E3,E4{E_1},{E_2},{E_3},{E_4} respectively. If L1{L_1} goes intoE1{E_1}, L2{L_2} goes intoE2{E_2}, L3{L_3}goes intoE3{E_3}, L4{L_4} goes into the envelope E4{E_4} then only we can say that all the letters go into the right envelopes and hence there is only one possible way of that.
Also we know that total number of ways in which four letters can be put into the four envelopes=4! = 4! =24 = 24
As we know that nCr=n!(n−r)!r!{}^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}
So 4C4=4!(4−4)!4!{}^4{C_4} = \dfrac{{4!}}{{(4 - 4)!4!}} and we also know that 0!=10! = 1
Hence we get that 4C4=4!0!4!=1{}^4{C_4} = \dfrac{{4!}}{{0!4!}} = 1
So there is only one possible way to put the 44 letters in 44 envelopes.
Also we know that total number of ways in which four letters can be put into the four envelopes =4! = 4! =24 = 24
So the probability of letter going in the right envelop is
P(E)=number of favourable outcomes total number of conditions=124P(E) = \dfrac{{{\text{number of favourable outcomes }}}}{{{\text{total number of conditions}}}} = \dfrac{1}{{24}}
But we want the probability of letter not going into the right envelope so that will be
1−P(E) ⇒1−124 ⇒2324  1 - P(E) \\\ \Rightarrow 1 - \dfrac{1}{{24}} \\\ \Rightarrow \dfrac{{23}}{{24}} \\\

Note:
nCr{}^n{C_r} refers to the number of ways in which r number of items can be chosen from n number of different items and also we know that nCr=nCn−r{}^n{C_r} = {}^n{C_{n - r}} and also nCr=n!(n−r)!r!{}^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}