So, lets say I have $4$ burglars, and I have a custom pistol with $10$ bullets in it. How many possible ways can I put $10$ bullets into EXACTLY $2$ burglars and let the other $2$ get away?
Note that I am NOT looking for an ANSWER, I am only asking if it is possible for someone to help me understand why my initial conditions are incorrect. In particular, can someone confirm if what I am doing below is correct?
*My technique is to use the inclusion exclusion formula where it is
$$E_2 = S_2 - \binom{3}{1}S_3 + \binom{4}{2}S_4$$
where $S_i$ stands for the possibilities where $i$ burglars get a bullet in them.
Now, lets say I want to find $S_2$, which is the number of ways to distribute $10$ bullets amongst $2$ burglars. Would the answer to this perhaps be
$$\binom{4}{2}(10^2 - 2)$$
where we first choose $2$ burglars out of the $4$ and then with the $2$ burglars each can have $10$ bullets so $10^2$ and then I minus $2$ because each has to have at least $1$.
Can someone please confirm if I did $S_2$ correctly?
The answer is $6132$ just in case anyone is wondering...