While it is not possible to find an exact answer, it is possible to find the best approximation. This is the method that many biologists use to estimate the population of a given animal in a given area. In fact, the animal analogy is very apt.
Imagine that a biologist randomly catches A animals and puts tags on all of them. Then, the biologist releases all of them. Assuming no animals breed, die, enter, or leave, the population will remain constant. Later, the biologist randomly catches B animals. This simulates the fact that John and David worked independently, since both catches were completely random and came from the same set. The biologist notices that of the B animals that he has just caught, C of them have the tags that he put on last time. The ratio of B to C is equal to the ratio of the total animal population (T) to A; in short, B/C=T/A. Ergo, T=(AB)/C. Normally, the biologist would be done here, but we have to do a little more manipulation.
Let's come back to the original problem: (AB)/C is the best estimate for the total number of errors. Between John and David, they caught A+B-C mistakes. The number of mistakes they did not catch would obviously be (AB)/C-(A+B-C). We could finish here.
However, further mathematical manipulation resolves to (AB-AC-BC+CC)/C. By factoring, we get (A(B-C)-C(B-C))/C, which factors again to our elegant answer, ((A-C)(B-C))/C.
Imagine that a biologist randomly catches A animals and puts tags on all of them. Then, the biologist releases all of them. Assuming no animals breed, die, enter, or leave, the population will remain constant. Later, the biologist randomly catches B animals. This simulates the fact that John and David worked independently, since both catches were completely random and came from the same set. The biologist notices that of the B animals that he has just caught, C of them have the tags that he put on last time. The ratio of B to C is equal to the ratio of the total animal population (T) to A; in short, B/C=T/A. Ergo, T=(AB)/C. Normally, the biologist would be done here, but we have to do a little more manipulation.
Let's come back to the original problem: (AB)/C is the best estimate for the total number of errors. Between John and David, they caught A+B-C mistakes. The number of mistakes they did not catch would obviously be (AB)/C-(A+B-C). We could finish here.
However, further mathematical manipulation resolves to (AB-AC-BC+CC)/C. By factoring, we get (A(B-C)-C(B-C))/C, which factors again to our elegant answer, ((A-C)(B-C))/C.
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