7 people have left their bags outside temple and after returning picked one bag each at random. In how many ways at least one and atmost three of them get their own bags?

Number of ways of getting one correct 

${=}^{\text{7}}{\text{C}}_{\text{1}}.6!(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!}){=}^{\text{7}}{\text{C}}_{\text{1}}.\left(265\right)$

Number of ways of getting two correct

${=}^{\text{7}}{\text{C}}_{\text{2}}.5!(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}){=}^{\text{7}}{\text{C}}_{\text{2}}.\left(44\right)$

Number of ways of getting three correct

${=}^{\text{7}}{\text{C}}_{\text{3}}.4!(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}){=}^{\text{7}}{\text{C}}_{\text{3}}.\left(9\right)$

Required number of ways =  ${ }^{\text{7}}{\text{C}}_{\text{3}}.9{+}^{\text{7}}{\text{C}}_{\text{2}}.44{+}^{\text{7}}{\text{C}}_{\text{1}}.265$

$\therefore \text{ Required number of ways }=315+924+1855=3094$