Question

Seven people leave their bags outside temple and while returning after worshiping the deity, picked one bag each at random. In how many ways at least one and at most three of them get their correct bags?
A. \[{}^7{C_3} \cdot 9 + {}^7{C_5} \cdot 44 + {}^7{C_1} \cdot 265\]
B. \[{}^7{C_6} \cdot 265 + {}^7{C_5} \cdot 9 + {}^7{C_7} \cdot 44\]
C. \[{}^7{C_5} \cdot 9 + {}^7{C_2} \cdot 44 + {}^7{C_1} \cdot 265\]
D. None of these

Answer 1

Hint: There will be three cases that need to be considered. First case one person can pick his correct bag. Second case two person can pick their correct bags. Third case three person pick their correct bags. Add the number of ways of the above case to get the correct answer.

Formula Used: The derangement formula for n objects is \[n!\left( {1 - \dfrac{1}{{1!}} + \dfrac{1}{{2!}} - \dfrac{1}{{3!}} + \cdots + \dfrac{1}{{n!}}} \right)\] when n is even. Or \[n!\left( {1 - \dfrac{1}{{1!}} + \dfrac{1}{{2!}} - \dfrac{1}{{3!}} + \cdots - \dfrac{1}{{n!}}} \right)\] when n is odd.

Complete step by step solution: There are three possible situations.
First case: only one person picks the correct bag and the rest are not.
Second case: Two persons pick the correct bag and the rest are not.
Third case: Three persons pick the correct bag and the rest are not.
First case:
The number of ways that one person picks the correct bag and the rest of the 6 are not is
\[{}^7{C_1} \times \]( derangement of 6 object)
\[ = {}^7{C_1} \times 6!\left( {1 - \dfrac{1}{{1!}} + \dfrac{1}{{2!}} - \dfrac{1}{{3!}} + \dfrac{1}{{4!}} - \dfrac{1}{{5!}} + \dfrac{1}{{6!}}} \right)\]
\[ = {}^7{C_1} \times 6!\left( {1 - 1 + \dfrac{1}{2} - \dfrac{1}{6} + \dfrac{1}{{24}} - \dfrac{1}{{120}} + \dfrac{1}{{720}}} \right)\]
\[ = {}^7{C_1} \times 6!\left( {\dfrac{{360 - 120 + 30 - 6 + 1}}{{720}}} \right)\]
\[ = {}^7{C_1} \times 6!\left( {\dfrac{{360 - 120 + 30 - 6 + 1}}{{720}}} \right)\]
\[ = {}^7{C_1} \cdot 6!\left( {\dfrac{{265}}{{720}}} \right)\]
\[ = {}^7{C_1} \times 265\]
second case:
The number of ways that two persons pick the correct bag and the rest of the 5 are not is
\[{}^7{C_2} \times \]( derangement of 5 object)
\[ = {}^7{C_2} \times 5!\left( {1 - \dfrac{1}{{1!}} + \dfrac{1}{{2!}} - \dfrac{1}{{3!}} + \dfrac{1}{{4!}} - \dfrac{1}{{5!}}} \right)\]
\[ = {}^7{C_2} \times 5!\left( {1 - 1 + \dfrac{1}{2} - \dfrac{1}{6} + \dfrac{1}{{24}} - \dfrac{1}{{120}}} \right)\]
\[ = {}^7{C_2} \times 5!\left( {\dfrac{{60 - 20 + 5 - 1}}{{5!}}} \right)\]
\[ = {}^7{C_2} \times 44\]

Third case:
The number of ways that three persons pick the correct bag and the rest of the 4 are not is
\[{}^7{C_3} \times \]( derangement of 4 object)
\[ = {}^7{C_3} \times 4!\left( {1 - \dfrac{1}{{1!}} + \dfrac{1}{{2!}} - \dfrac{1}{{3!}} + \dfrac{1}{{4!}}} \right)\]
\[ = {}^7{C_3} \times 4!\left( {1 - 1 + \dfrac{1}{2} - \dfrac{1}{6} + \dfrac{1}{{24}}} \right)\]
\[ = {}^7{C_3} \times 4!\left( {\dfrac{{12 - 4 + 1}}{{24}}} \right)\]
\[ = {}^7{C_3} \times 9\]
The total number of ways at least one and at most three of them get their correct bags is
\[ = {}^7{C_1} \times 265 + {}^7{C_2} \times 44 + {}^7{C_3} \times 9\]

Option ‘A’ is correct

Note: Students often do mistake to find the ways. They did not multiply derangement with the combination. Here we have to multiply derangement.