Past exam of the year 1 of the computer science course of the University of Oxford / 2013 / Discrete mathematics, probability and continuous mathematics / 3 / c / Solution

For finite sets, the inclusion-exclusion principle statesAn element present in $r$ of the sets is counted ${\sum }_{k=1}^r(-1{)}^{k+1}(\genfrac{}{}{0ex}{}{r}{k})=1$ times, which explains the alternating corrections.

There are $\lfloor 999/d\rfloor$ positive integers strictly below $1000$ divisible by $d$. For the divisors $2,5,7$, the single counts are $499,199,142$. Their pairwise intersections correspond to divisibility by $10,14,35$, giving $99,71,28$. The triple intersection has $\lfloor 999/70\rfloor =14$ elements. Therefore