Question

A person writes letters to his $5$ friends and addresses the corresponding envelopes. Number of ways in which the letters can be placed in the envelope, so that at least two of them are in the wrong envelopes?

Answer 1

Hint:At least two of them are in wrong envelopes means minimum two letters are in the wrong envelopes and maximum five letters are in the wrong envelope. Use the formula for number of ways in which at least two of them are in the wrong envelopes:
$\sum\limits_{r = 2}^5 {{}^n} {C_{n - r}}{D_r}$, here ${D_r} = r!\left[ {1 - \dfrac{1}{{1!}} + \dfrac{1}{{2!}}....\dfrac{1}{{r!}}} \right]$

Complete step-by-step answer:
We are given that a person writes letters to his $5$ friends and addresses the corresponding envelopes.
We have to evaluate the number of ways in which the letters can be placed in the envelope, so that at least two of them are in the wrong envelopes.
For this we use the following formula:
$\sum\limits_{r = 2}^5 {{}^n} {C_{n - r}}{D_r}$, here ${D_r} = r!\left[ {1 - \dfrac{1}{{1!}} + \dfrac{1}{{2!}}....\dfrac{1}{{r!}}} \right]$
The range of the summation is from $2$to $5$because at least two of them are in wrong envelopes means minimum two letters are in the wrong envelopes and maximum five letters are in the wrong envelope.
Expand the summation.
\[
   \Rightarrow {}^5{C_{5 - 2}}{D_2} + {}^5{C_{5 - 3}}{D_3} + {}^5{C_{5 - 4}}{D_4} + {}^5{C_{5 - 5}}{D_5} \\
   = {}^5{C_3}{D_2} + {}^5{C_2}{D_3} + {}^5{C_1}{D_4} + {}^5{C_0}{D_5} \\
 \]
Use the formula ${}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$ to solve the expression.
\[ \Rightarrow \dfrac{{5!}}{{3!2!}}2!\left[ {1 - \dfrac{1}{{1!}} + \dfrac{1}{{2!}}} \right] + \dfrac{{5!}}{{3!2!}}3!\left[ {1 - \dfrac{1}{{1!}} + \dfrac{1}{{2!}} - \dfrac{1}{{3!}}} \right] + \dfrac{{5!}}{{1!4!}}4!\left[ {1 - \dfrac{1}{{1!}} + \dfrac{1}{{2!}} - \dfrac{1}{{3!}} + \dfrac{1}{{4!}}} \right] + \dfrac{{5!}}{{0!5!}}5!\left[ {1 - \dfrac{1}{{1!}} + \dfrac{1}{{2!}} - \dfrac{1}{{3!}} + \dfrac{1}{{4!}} - \dfrac{1}{{5!}}} \right]\]
Use the formula $n! = n(n - 1)(n - 1)....3 \times 2 \times 1$to solve the above expression.
\[
   \Rightarrow \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{3 \times 2 \times 1 \times 2 \times 1}}2 \times 1\left[ {1 - \dfrac{1}{1} + \dfrac{1}{{2 \times 1}}} \right] + \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{3 \times 2 \times 1 \times 2 \times 1}}3 \times 2 \times 1\left[ {1 - \dfrac{1}{1} + \dfrac{1}{{2 \times 1}} - \dfrac{1}{{3 \times 2 \times 1}}} \right] \\
   + \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{4 \times 3 \times 2 \times 1}}4 \times 3 \times 2 \times 1\left[ {1 - \dfrac{1}{1} + \dfrac{1}{{2 \times 1}} - \dfrac{1}{{3 \times 2 \times 1}} + \dfrac{1}{{4 \times 3 \times 2 \times 1}}} \right] + \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{5 \times 4 \times 3 \times 2 \times 1}}5 \times 4 \times 3 \times 2 \times 1\left[ {1 - \dfrac{1}{1} + \dfrac{1}{{2 \times 1}} - \dfrac{1}{{3 \times 2 \times 1}} + \dfrac{1}{{4 \times 3 \times 2 \times 1}} - \dfrac{1}{{5 \times 4 \times 3 \times 2 \times 1}}} \right] \\
 \]
$ = 20\left[ {\dfrac{1}{2}} \right] + 60\left[ {\dfrac{1}{2} - \dfrac{1}{6}} \right] + 120\left[ {\dfrac{1}{2} - \dfrac{1}{6} + \dfrac{1}{{24}}} \right] + 120\left[ {\dfrac{1}{2} - \dfrac{1}{6} + \dfrac{1}{{24}} - \dfrac{1}{{120}}} \right]$
\[
   = 10 + 60\left[ {\dfrac{{3 - 1}}{6}} \right] + 120\left[ {\dfrac{{12 - 4 + 1}}{{24}}} \right] + 120\left[ {\dfrac{{60 - 20 + 5 - 1}}{{120}}} \right] \\
   = 10 + 20 + 45 + 44 \\
   = 119 \\
 \]
Hence, Number of ways in which the letters can be placed in the envelope, so that at least two of them are in the wrong envelopes is $119$.

Note:
 The difference between at least and at most is:
At least $2$ events mean the minimum number of events are $2$ and at most $2$ events means maximum number of events are $2$.