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Q. Numbers $1$ to $1000$ which are divisible by $60$ but not by $24$ is

Permutations and Combinations

Solution:

Let $n(A)=$ Number of divisible by 60
$=(60,120, \ldots . ., 960)=16$
$n(B)=$ Number of divisible by $24=(24,48 m \ldots ., 984)$
$=41$
$n(A \cap B)=$ Number divisible by both
$=120+240+\ldots .+960=8$
Hence, $n(A \cap \bar{B})=n(A)-n(A \cap B)$
$=16-8=8$