<0MB1N4T10N5

	5 men	4 women
	2 married couples
	committee of 2 men and 2 women without any married couples
[if you don't have time to read this, at least look at SOLUTION 3]

SOLUTION 1:
Let:
M1 = married man 1	M2 = married man 2
W1 = married woman 1	W2 = married woman 2
U = unmarried opposite sex
[assume no married persons unless stated]

U U	+	M1 U	+	M1 W2	+	M2 U	+	M2 W1	+	U W1	+	U W2	+	M1M2 U	+	W1W2 U
3C2*2C2	+	3C1*2C2	+	3C1*2C1	+	3C1*2C2	+	3C1*2C1	+	3C2*2C1	+	3C2*2C1	+	1*2C2	+	1*3C2
37


SOLUTION 2:
no married persons:					3C2*2C2
one married man + women excluding his wife * 2:		3C1*3C2*2
one married women + no married men * 2:			2C1*3C2*2
both married men + no married women:			1*2C2
both married women + no married men:			1*3C2
sum: 							37


SOLUTION 3:
total combinations (including married persons):		5C2*4C2
both married couples:					1
married couple but not all of other couple * 2:		[ 4C1*3C1 - 1 ]*2
total - married persons combinations:			37

4C1*3C1 represents all combinations of remaining men and women for the remaining 2 seats.
- 1 to take away the combination of having both married couples


~~for you Simon; ace that stats paper!
Marko