Imagine I have 76 million boxes, and I want to put only 2 particles maximum for each box, I want to find the total numbers of ways (combinations) that involves placing 2 particle in each box given that you have 40,000 particles in total.
I have attempted the question, and Here is my step:
$$\Sigma_{i=1}^{20000} \frac{76000000!}{(2!^{i}1!^{N-i})}$$
Eventually, I would divide this number by 76000000! to compute the probability of placing 2 particles in each box.
To compute the higher order terms, I would also start including the case when some boxes have 3 particles, but the majority has 2 particles.
To convince you that my initial thoughts is correct: Let's imagine a simpler problem with 4 boxes, and only 3 particles.
If you draw all the possible cases with 2 particles maximum for each box you realise placing the first two particles in the same box, you have 4 choices, after that you are left with 3 choices for placing only one particle. So the number of ways for 2 particles maximum in a box is Number of ways = 4!/(2!1!)