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

The correct option is A $2\cdot 7\cdot 13\cdot 17$
Number of ways only one get his bag
$={}^7{C}_{1}6!(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!})={}^7{C}_1\cdot 265$
Number of ways only two get his bag
$={}^7{C}_{2}5!(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!})={}^7{C}_2\cdot 44$
Number of ways only three get his bag
$={}^7{C}_{3}4!(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!})={}^7{C}_3\cdot 9$
So required number of ways
$=2\cdot 7\cdot 13\cdot 17$
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