Question

The number of ways in which 7 letters can be put in 7 envelopes such that exactly four letters are in wrong envelopes is :
A) 300
B) 315
C) 325
D) 10355

Answer 1

First of all we will select 3 letters as here in the Question it is mentioned that 3 letters should be correct envelopes. So, selection of 3 -letters out of 7 letters is$^{7}{C_3}$. Now we have left with 4 letters with 9 envelope. Because 3-letters already left inside 3 envelope in a correct sequence. So, According to the given question 4 letters should be arrange in 7 envelope which is equal to 4! $\left[ {1 - \dfrac{1}{{1!}} + \dfrac{1}{{2!}} - \dfrac{1}{{3!}} + \dfrac{1}{{4!}}} \right]$

Complete step by step solution: First of all selection of 3 letters out of 7 letters will be ${e_2},\;{l_1}$
Then, we have left with 4 envelope and 4 - letters.
So, the ways of dearrangement is :-
$4!\left[ {1 - \dfrac{1}{{1!}} + \dfrac{1}{{2!}} - \dfrac{1}{{3!}} + \dfrac{1}{{4!}}} \right]$
$4!\left[ {\dfrac{1}{2} - \dfrac{1}{6} + \dfrac{1}{{24}}} \right]$
$4!\left[ {\dfrac{{12 - 4 + 1}}{{24}}} \right]$
$4!\left[ {\dfrac{9}{4}} \right]$
$\left( {4 \times 3 \times 2 \times \dfrac{9}{{24}}} \right) = 9$
And arrangement of 3 letters out of 7letters are $7{C_3}$
$= \left( {\dfrac{{7!}}{{4!\, \times 3!}}} \right) = \left( {\dfrac{{\left( {7 \times 6 \times 5 \times 4 \times 3} \right) \times 2}}{{\left( {4 \times 3 \times 2} \right) \times \left( {2 \times 2} \right)}}} \right) = (35)ways.$
So, the no. of ways that 3 letters correctly emery into correct envelope is 35 ways.
And no. of ways that 4 letters dearrange 4 – envelopes are: 9 ways.
So, to arrange 3 letters in the correct envelope and dearrange 4 letters in 4 wrong envelopes are :
= (35 × 9)
= (315) ways total.

Hence option (B) is correct.

Note: We must know first about the dearrangement theorem, we have n – letters and n envelopes, and we don’t want the letters should go inside the corrected envelope like ${l_1}$ should not go inside ${e_1},\;{l_2}$should not go inside ${e_2}$ and so, on. If - ${l_1}$ represents letter 1 and e1 represents envelope 1.
So, The ways of dearrangement :
$n!\left[ {1\dfrac{{ - 1}}{{1!}} + \dfrac{1}{{2!}} + - - - - {{\left( { - 1} \right)}^n}\dfrac{1}{{n!}}} \right]$
And also selection of 3 letter out of 7 letters is a combination. Which is equal to $\left( {^{7}{C_3}} \right) = \left( {\dfrac{{7!}}{{\left( {7 - 3} \right)!\, \times 3!}}} \right)$
So, by the help of these 2 – concepts we can find its Solution.