Probability problem of 220 people randomly selecting only 12 of 35 exclusive options. There are 220 people and 35 boxes filled with trinkets.
Each person takes one trinket out of a random box. 
What is the probability that the 220 people will have grabbed a trinket from exactly 12 different boxes?
I'm trying to calculate the probability of grabbing a trinket from at most 12 boxes, $P(12)$. Then calculate $P(11)$ with the answer being $P(12)-P(11)$ but I'm drawing blank.
 $P(12) = 1-(23/35)^{220}$ doesn't look right to me.

If the probability that the first $n$ people have chosen from exactly $c$ boxes out of a possible $t$ total boxes [$t=35$ in this case] is $p(n,c,t)$ then $$p(n,c,t)=\frac{c \times p(n-1,c,t)+(t-c+1)\times p(n-1,c-1,t)}{t}$$ starting with $p(0,c,t)=0$ and $p(n,0,t)=0$ except $p(0,0,t)=1$.
Using this gives $p(220,12,35) \approx 4.42899922 \times 10^{-94}$. This is close to but not exactly the naive ${35 \choose 12}\times \left(\frac{12}{35}\right)^{220} \approx 4.42899948\times 10^{-94}$.
