A bag contains 6 blue and 6 green balls. Pairs of balls are drawn without replacement until the bag is empty. The probability that each drawn pair consists of one blue and one green ball is :

The number of ways to divide 12 objects into 6 unordered pairs is given by the formula for partitions into equal-sized groups:

$$n(S) = \frac{12!}{(2!)^6 \times 6!}$$

For each of the 6 pairs to have exactly one blue ball and one green ball, we must pair each of the 6 unique blue balls with one of the 6 unique green balls.

The first blue ball has 6 choices of green balls to pair with.

The second blue ball has 5 choices of remaining green balls.

The third blue ball has 4 choices, and so on.

$$n(E) = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$$

$$\text{Probability} = \frac{n(E)}{n(S)} = \frac{6!}{\frac{12!}{2^6 \times 6!}} = \frac{(6! \times 6!) \times 2^6}{12!}$$

$$\text{Probability} = \frac{16}{231}$$