Question

Seven people leave their bags outside a temple and returning after worshiping picked one bag each at random.In how many ways atleast one and atmost three of them get their correct bags?

• A. $3094$
• B. $3095$
• C. $3096$
• D. $3093$

Answer 1

The correct option is A $3094$

Number of ways in which one get his own bag
${}^7{C}_{1}6!(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!}){=}^7{C}_1\times 265$
Number of ways in which two get his own bag
${}^7{C}_{2}5!(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}){=}^7{C}_2\times 44$
Number of ways in which three get his own bag
${}^7{C}_{3}4!(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}){=}^7{C}_3\times 9$
$\therefore$ Required number of ways
= ${}^7{C}_3\times 9{+}^7{C}_2\times 44{+}^7{C}_1\times 265=3094$