There were a number of bats in a cave. Two bats see out of the right eye, three could see out of the left eye, four could not see out of the left eye, and five could not see out of the right eye. What is the least number of bats in the cave and what might their eyesight have been?

To solve this problem, we need to find the least number of bats in the cave and determine the different configurations of their eyesight.

Let's break down the information given:

- Two bats see out of the right eye.
- Three bats see out of the left eye.
- Four bats cannot see out of the left eye.
- Five bats cannot see out of the right eye.

To find the least number of bats, we need to consider the bats with the most restricted eyesight. In this case, we have four bats that cannot see out of the left eye and five bats that cannot see out of the right eye. Since these two sets of bats have no overlap, we can assume that all four bats with restricted left eyesight are also part of the group with restricted right eyesight, making a total of four bats.

Now, let's determine the remaining bats. Since we already have four bats with restricted eyesight, we subtract this number from the total number of bats mentioned in the problem. Mathematically, if we represent the total number of bats as 'x', we can deduce that:

x - 4 = 2 (bats with unrestricted right eyesight) + 3 (bats with unrestricted left eyesight)

Combining like terms, we get:

x - 4 = 5

Solving for 'x', we find:

x = 9

Therefore, the least number of bats in the cave is 9. Now let's determine the different eye configurations:

- Two bats see out of the right eye.
- Three bats see out of the left eye.
- Four bats cannot see out of the left eye.
- Five bats cannot see out of the right eye.

In summary, there are 9 bats in the cave, with two bats having unrestricted eyesight in the right eye, three bats having unrestricted eyesight in the left eye, four bats having restricted eyesight in both eyes, and the remaining five bats having restricted eyesight in the right eye only.