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?

  • $\begingroup$ Please, type such short texts. It takes 5 minutes... $\endgroup$
    – Jean Marie
    Sep 25, 2020 at 10:09
  • $\begingroup$ Your first attempt counts each committee three times, once for each way you could have designated two of the three people from each city (or county) as the two people from that locale. $\endgroup$ Sep 25, 2020 at 10:21

2 Answers 2


You have two selection rounds. In the first round, you select two people from each group. In the second round you select one people from any group.

Consider a group of which 3 members are chosen; A, B, and C. Any of these 3 can be chosen in the second round, which is why your answer is three - fold the correct answer.

  • $\begingroup$ But don't we remove the possibility of items being chosen for a second time by only choosing from 12 and not 18? $\endgroup$
    – PaulBaul11
    Sep 25, 2020 at 16:13
  • 1
    $\begingroup$ That is not what I mean. What i mean is, with your method, we can choose A and B in the first round then C in the second, A and C in the first then B in the second, or B and C in the first then A in the second. Three possibilities yielding the same committee; A,B,C. $\endgroup$ Sep 25, 2020 at 16:23

Your calculation will have duplicate counts - in fact exactly $2$ times more.

One simple way to solve the problem is to see that -
Selecting $7$ people for the committee with min. $2$ people from each city means $2$ members each from two cities and $3$ members from the third city.

So, number of ways = $ {5 \choose 3} \times {6 \choose 2} \times {7 \choose 2} + {5 \choose 2} \times {6 \choose 3} \times {7 \choose 2} + {5 \choose 2} \times {6 \choose 2} \times {7 \choose 3} = 12600$

Which is exactly $3$ times of your calculation ($2$ times more). So the other way is to just calculate the way you did, understand the overcounting and divide by $3$.

  • $\begingroup$ @PaulBall11 It is worth noting that Math Lover's approach critically depends on the idea that he has split the # of satisfying choices into 3 mutually exclusive groups. That is, each group corresponds to which city will contribute 3 people. It is absolutely impossible for a satisfying choice to belong to more than one of the 3 groups. That is, if a given choice involves Gloucester contributing 3 people, it is impossible for the same choice to also involve one of the other two cities contributing 3 people. It is this mutual exclusivity approach that prevents over counting. $\endgroup$ Sep 25, 2020 at 11:52

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