Counting (Combinations) Suppose there are 10 people apart of a club: A, B, C, D, E, F, G, H, I, and J. They decide to go to a restaurant for a club outing, but there isn't one table to seat all of them, so they decide to take one table that seats four people and two tables that seat three people. Based on this, what is the probability that Person G and Person J sit at the same table?


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*Ways for both Person G and Person J to both sit at the 4-person table: 6 = (4 choose 2)

*Ways for both Person G and Person J to both sit at the 3-person table: 3 = (3 choose 2)

*Ways for both Person G and Person J to both sit at the other 3-person table: 3 = (3 choose 2)


Total of $6+3+3=12$ ways for Person G and Person J to both sit at the same table.
I've determined the number of ways that Person G and Person J can sit together at each table, but I don't know how to properly count the total different ways (i.e., the denominator) to determine the probability.

I am considering the tables to be identical because you haven, t clarified that in the question, so in question we have 6 seats in two tables, where each  table contains exact three seats, and one table which contain exact 4 seats. 
So we treat both ,G, J as together one. 
Case 1=when both seat on table 1 consist of 4 people 
So we have 8 options left to choose 2 spots because rest 2 were already occupied by them. , and similarly for next table we have only 6 options left to choose three people and left 3 people have only one choice that is last, and for cases 2)you can count the similar way. 
Now total umber of possible cases are we choose 4 out of 10 people from table one, and 3 out of 6 people, and left three with left chairs. So by calculating the value you will get the answer $4/15$.
