Question

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?

• A. $2\cdot 7\cdot 13\cdot 17$
• B. $4\cdot 7\cdot 13\cdot 17$
• C. $7\cdot 13\cdot 17$
• D. $3\cdot 7\cdot 13\cdot 17$

Answer 1

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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