7 people have 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 correct bags

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$^{7} C_{6} (265) +^{7} C_{5} (9) +^{7} C_{4} (44)$

$7 C_{1} (265) + 7 C_{2} (44) + 7 C_{3} (9)$

$^{7} C_{6} (9) +^{7} C_{5} (44) +^{7} C_{1} (265)$

$^{7} C_{3} (9) -^{7} C_{2} (44) -^{7} C_{1} (265)$

answer is C.

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Detailed Solution

No. of ways of getting one correct $= 7 C_{1} 6! (1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \frac{1}{6!}) = 7 C_{1} (265)$
No. of ways of getting two correct $= 7 C_{2} 5! (1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!}) = 7 C_{2} (44)$
No. of ways of getting three correct $= 7 C_{2} 4! (1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!}) = 7 C_{3} (9)$
Required no. of ways $= 7 C_{1} (265) + 7 C_{2} (44) + 7 C_{3} (9)$