Four dice are rolled once. The number of possible outcomes in which at least one dice shows $3$ is

1296

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625

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671

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none of these

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Solution

The correct option is C $671$
$N o . o f w a y s i n w h i c h a n y n u m b e r a p p e a r i n g i n o n e d i c e = 6 N o . o f w a y s i n w h i c h 2 a p p e a r i n g o n e d i c e = 1 N o . o f w a y s i n w h i c h 2 d o e s n o t a p p e a r i n o n e d i c e = 5 T h e r e a r e 4 d i c e . G e t t i n g 2 i n a t l e a s t o n e d i c e = G e t t i n g a n y n u m b e r i n a l l t h e 4 d i c e - G e t t i n g n o t 2 i n a n y o f t h e 4 d i c e . = (6 \times 6 \times 6 \times 6) - (5 \times 5 \times \times 5 \times 5) = 1296 - 625 = 671$

Hence, Option $C$ is correct answer.

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