Question

A post man has to deliver five letters to five different houses. Mischievously, he posts one letter through each door without looking to see if it is the correct address. In how many different ways could he do this so that exactly two of the five houses receive the correct letters?

Answer 1

In this particular type of question assume A, B, C, D and E are the house numbers then consider all possible cases such that two of the houses can get the correct letters so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Let us consider the five houses are A, B, C, D and E.
Now postmen post five letters in these houses without looking at the address such that exactly two houses can get correct letters.
So the possible number of cases are:
When the correct houses are A and B, the number of arrangements of the remaining (5 – 2) = 3, houses is 3!
In these arrangements there is only one case in which all of the houses can get the correct letters so we have to subtract that case.
And there is also a case when one of them can get the correct letter and two of them can get the incorrect letters, so in this case there are 3 possibilities.
so the total possibilities we have to subtract from 3! is (1 + 3) = 4
So the possible arrangements when A and B are correct houses are (3! – 4) = (3.2.1) – 4 = (6 – 4) = 2.
Similarly for rest of the cases when (A and C) are correct houses, (A and D) are correct houses, (A and E) are correct houses, (B and C) are correct houses, (B and D) are correct houses, (B and E) are correct houses, (C and D) are correct houses, (C and E) are correct houses, (D and E) are correct houses.
So the possible cases are 10 cases including the first case (i.e. A and B).
So the total number of arrangements are such that exactly two houses can get correct letters is the product of the number of cases and the number of arrangements of any particular case.
So the total number of arrangements are such that exactly two houses can get correct letters is
= 10(2) = 20.
So this is the required answer.

Note: Whenever we face such types of questions the key concept we have to remember is that the number of ways to arrange n different objects is (n!), and in these cases if there are (r) number of cases we do not want, so we have to subtract these cases from the number of cases so the resultant arrangement is (n! – r) and if there are q such possible cases that arise then the number of arrangements is q times (n! – r).