The number of ways of selecting 15 teams from 15 men and 15 women, such that each team consists of a man and a woman is

We have 15 distinct men, which we may label as $$M_1,\,M_2,\,\dots ,\,M_{15}$$, and 15 distinct women, which we may label as $$W_1,\,W_2,\,\dots ,\,W_{15}$$. We must form 15 teams, with each team containing exactly one man and exactly one woman, and every person must be used exactly once. In other words, we must pair each man with a unique woman.

To count the total number of possible pairings, we proceed man by man and apply the basic rule of counting (also called the multiplication principle):

For the first man $$M_1$$, any of the 15 women can be chosen, giving $$15$$ possibilities.

After we fix the woman for $$M_1$$, there remain $$14$$ women for the second man $$M_2$$, giving $$14$$ possibilities.

Continuing in the same fashion, for $$M_3$$ there are $$13$$ available women, for $$M_4$$ there are $$12$$ women, and so on, until the very last man $$M_{15}$$, who will have exactly $$1$$ woman left.

By the multiplication principle, we multiply all these successive counts:

$$ 15 \times 14 \times 13 \times \dots \times 2 \times 1 $$

This product is precisely the factorial of 15, which is written as $$15!$$. No other choices or arrangements remain to be made, so this product accounts for every possible way to create 15 distinct man-woman teams using every person exactly once.

Hence, the total number of ways is

$$ 15! $$

Therefore, the correct option among the choices given is Option B.

Hence, the correct answer is Option B.