Explanation:

Let the husbands be h1, h2, h3…….h9 and wives be w1, w2, w3…….w9 Choosing 2 husband=⁹C₂=36

Choosing 2 wives = ⁷C₂ =21 (Here we should have taken ⁹C₂, but as per the question no husband and wife can play in the same game, so we eliminated 2 wives of chosen husbands)

We know that in mixed double game, it is played as m1w1 vs m2w2.

So, out of 36 men and 21 women two arrangement is possible, either m1w1 vs m2w2 or m2w1 vs m1w2.

Therefore, total ways = 36 × 21 × 2 = 1512 ways

Hence, option (C) is correct.