1. let the 5 couples be represented as

A(H,W); B(H,W); C(H,W); D(H,W); E(H,W); where H represent Husband and W wife.

now out of these 5 grps, u have to select 3; only then u will get a group of mutually unmarried ppl; so that can be done in 5C3 ways. However from within a grp (A,b,c,d e) you can select either H or W. and that can be done in 2C1 ways.

SO the total no of ways = 5C3 * 2C1 = 10 * 2 = 20.

2. its a question of no of permutations of 9 things where 3 are of one type, 4 of the other and 2 of the third type and that is 9! / (3! * 4! * 2!) .... a huge number, not worth the type spending on calculation...