In a town of 10,000 families it was found that $40\%$ family buy newspaper A, $20\%$ buy newspaper B and $10\%$ families buy newspaper C, $5\%$ families buy A and B, $3\%$ buy B and C and $4\%$ buy A and C. If $2\%$ families buy all the three newspapers, then number of families which buy A only is

$\mathrm{n}\left(\mathrm{A}\right)=40\% \mathrm{of} 10,000=4,000$

$\mathrm{n}\left(\mathrm{B}\right)=20\% \mathrm{of} 10,000=2,000$

$\mathrm{n}\left(\mathrm{C}\right)=10\% \mathrm{of} 10,000=1,000$

$\mathrm{n}(\mathrm{A}\cap \mathrm{B})=5\% \mathrm{of} 10,000=500$

$\mathrm{n}(\mathrm{B}\cap \mathrm{C}) =3\% \mathrm{of} 10,000=300$

$\mathrm{n}(\mathrm{C} \cap \mathrm{A})=4\% \mathrm{of} 10,000=400$

$\mathrm{n}(\mathrm{A}\cap \mathrm{B}\cap \mathrm{C})=2\% \mathrm{of} 10,000=200$

We want to find $\mathrm{n}(\mathrm{A}\cap {\mathrm{B}}^{\mathrm{c}}\cap {\mathrm{C}}^{\mathrm{c}})=\mathrm{n}[\mathrm{A}\cap {(\mathrm{B}\cup \mathrm{C})}^{\mathrm{c}}]$

$= \mathrm{n}\left(\mathrm{A}\right)– \mathrm{n}[\mathrm{A}\cap (\mathrm{B}\cup \mathrm{C}\left)\right]=\mathrm{n}\left(\mathrm{A}\right)–\mathrm{n}\left[\right(\mathrm{A}\cap \mathrm{B})\cup (\mathrm{A}\cap \mathrm{C}\left)\right]$

$=\mathrm{n}\left(\mathrm{A}\right)–\left[\mathrm{n}\right(\mathrm{A}\cap \mathrm{B})+\mathrm{n}(\mathrm{A}\cap \mathrm{C})–\mathrm{n}(\mathrm{A}\cap \mathrm{B}\cap \mathrm{C}\left)\right]$

$=400-\left[500+400-200\right]=4000-700=3300.$