Probability that two distinguishable objects to be included Say we are distributing 19 distinguishable meals to $7$ distinguishable people, with possible repetition, uniformly at random.  I am assuming $|\Omega|=19^7$.  I am trying to find the probability that two arbitrary meals $a$ and $b$ are both distributed.
I originally thought it would be $\frac{7\cdot 6\cdot 19^5}{19^7}$ but I think I am overcounting...
Could it be (edited): $\frac{\sum_{i=1}^6\sum_{j=1}^{7-i}\binom{7}{i}\binom{7-i}{j}17^{7-i-j}}{19^7}\approx 0.0892414$?  That looks way too complicated.  My thinking is that $i$ represents the number of $a$ meals distributed and $j$ represents the number of $b$ meals distributed (they both must be $\geq$ 1).
 A: I agreed with your last solution, using another method:
Let's again use the complement argument:
What are the ways that neither $a$ nor $b$ are chosen? 
Number of ways $a$ not chosen: $\dfrac{18^7}{19^7}$
Number of ways $b$ not chosen: $\dfrac{18^7}{19^7}$
However, we are counting the event that neither $a$ nor $b$ is chosen twice, so we must deduct $\dfrac{17^7}{19^7}$
So we get:
$$1-\bigg(\dfrac{18^7}{19^7}+\dfrac{18^7}{19^7}-\dfrac{17^7}{19^7}\bigg)=0.08924$$
I think my old answer is incorrect since I am not taking into account the fact that the people are distinguishable, but I'm not sure. If anyone else can confirm this, please do.
OLD ANSWER:
Let $A$ = event that $a$ is chosen and $B$ = event that $b$ is chosen
$$Pr(A)=1-Pr(A^c)=1-\bigg(\dfrac{18}{19}\bigg)^7$$
The last term means that all 7 people must choose something other than meal $a$
$$Pr(B|A)=1-Pr(B^c|A)=1-\bigg(\dfrac{18}{19}\bigg)^6$$
The power in this case is 6 since the first dish is already catered for in the first equation
$$Pr(A, B)=Pr(A)Pr(B|A)=\bigg(1-\bigg(\dfrac{18}{19}\bigg)^7\bigg)\bigg(1-\bigg(\dfrac{18}{19}\bigg)^6\bigg)=0.0873$$
