2)Hi!!!
Here goes the solution>>>
No. of people attending the draw = 2z
Thus, total no. of ways of selecting m winners = ^2zC_m.

It is known that,
if only one couple is among the m winners,
the couple can be chosen in ^zC₁ = z ways.

Clearly,
remaining m-2 winners are to be chosen from z-1 couples such that no 2 winners are from the same couple.

Now,
no. of ways of selecting the first winner = 2(z-1).
{becoz the winner can be either the husband or the wife}

Similarly,
no. of ways of selecting the second winner = 2(z-2)
.
.
.
no. of ways of selecting the (m-2)th winner = 2(z-(m-2))

Thus,
total no. of ways of selecting the m winners such that they include a couple also = (z).{2(z-1)}.{2(z-2)}...{2(z-(m-2))}
= (z).{2^m-2}.{^z-1P_m-2}

Clearly,
required probability = (z).{2^m-2}.{^z-1P_m-2} / (^2zC_m).