Question: A committee of $7$ people is to be chosen at random from $18$ volunteers.
The $18$ volunteers consist of $5$ people from Gloucester, $6$ from Hereford and $7$ from Worchester. The committee is to be chosen randomly.
Find the probability that the committee will include at least $2$ people from each of the three cities.
Attempt: First, I calculated the total number of ways of selecting the committee which I got as $^{18}C_{7} = 31824.$
Since we require at least $2$ people from each city, I calculated the number of ways of selecting exactly two people from each city which I got as $^{5}C_{2} \cdot ^{6}C_{2} \cdot ^{7}C_{2},$ and since we then require $1$ extra person whom can be from anywhere, I multiplied this by $^{12}C_{1}$ as there are $12$ people remaining after choosing $6$ and we only need $1$ more to get the full $7.$
This gives $37800,$ which is greater than the total number of ways to select the committee so I know this is wrong, but I don't understand what oversight I have made. I have tried another method where you sum the different combinations instead and I got the correct answer, but I don't understand why this method doesn't work.
Where have I gone wrong?