Let $m_1, ..., m_r$ be pairwise coprime numbers and $N=\prod\limits_{i=1}^r m_i$ . So I'm trying to calculate the proportion of the numbers 1 to N that are not divisible by any of the $m_i$, so I've used the Inclusion-Exclusion Principle, but I have a really long expression of sums and was wondering how I can simplify it?
So for the size of the set of all numbers between 1 and n not divisible by any of the $m_i$ is : $$N-\sum_{i} N/m_i + \sum_{i<j} N/m_im_j - \sum_{i<j<k} N/m_im_jm_k \quad + ...$$ Then for the proportion just dividing through by N I got: $$1-\sum_{i} 1/m_i + \sum_{i<j} 1/m_im_j - \sum_{i<j<k} 1/m_im_jm_k \quad + ...$$
So firstly I would suggest not dividing through by N, at least not initially. Imagine you have $r=3$, then $$ N-\sum_i \frac{N}{m_i} + ... = m_1m_2m_3 + m_1m_2 + m_1m_3 + m_2m_3 - m_1 - m_2 - m_3 - 1 $$ Then, clearly $$ m_1m_2m_3 + m_1m_2 + m_1m_3 + m_2m_3 - m_1 - m_2 - m_3 - 1 = (m_1-1)(m_2-1)(m_3-1) $$ This can be expanded to larger values of r.