Answer: $\mathbf A$
Explanation:
This can be solved using the inclusion-exclusion principle and Demorgan's Law:
$|\bar {A}| \cap \bar {|B|}\cap \bar {|C|} = 1000 - |A\cap B\cap C|= 1000 - (|A| + |B|+|C|) + (|A|\cap |
B|) + (|B|\cap |
C|) + |C|\cap |A|) - |A\cap B\cap C|\tag{1}$
$|A| = \Big\lfloor {\frac{1000}{17}}\Big \rfloor = 58 $
$|B| = \Big\lfloor {\frac{1000}{19}}\Big \rfloor = 52$
$|C| = \Big\lfloor {\frac{1000}{23}}\Big \rfloor = 43$
$|A\cap B| = \Big\lfloor {\frac{1000}{17.19}}\Big \rfloor = 3 $
$|B\cap C| = \Big\lfloor {\frac{1000}{19.23}}\Big \rfloor = 2$
$|C\cap A| = \Big\lfloor {\frac{1000}{23.17}}\Big \rfloor = 2$
$|A\cap B \cap C| = \Big\lfloor {\frac{1000}{17.19.23}}\Big \rfloor = 0$
On substituting these values in $\mathrm {{eq}^n} \; (1)$, we get:
Answer $=1000 -( 58+52+43)+3+2+2 = 854$
$\therefore \mathbf A$ is the correct option.