10 ( different ) people have the option of traveling to 20 ( different ) destination . On how many ways they can leave if we know that one can visit more destination ( the order of visits is not important ) , no one will visit more than three destinations , and that some people may give up the trip ?
Thank you in advance.
Based on the comments, I interpret the question to be asking in how many different ways $20$ destinations can be assigned to $10$ people such that no person is assigned more than $3$ destinations. This is a case of balls in bins with limited capacity, which can be solved using inclusion-exclusion, and the answer is
$$ \sum_{t=0}^{10}(-1)^t\binom{10}t\binom{10+20-4t-1}{10-1}=\sum_{t=0}^{10}(-1)^t\binom{10}t\binom{29-4t}{9}=44803 $$
(where, contrary to the standard convention, the binomial coefficients are taken to vanish when the upper argument is negative).