I highly doubt it but would there happen to be anyone out there nice enough to help me with this probability question? I would be ETERNALLY grateful if there was :D :D :D I've been staring at it for the past half hour and I haven't gotten anywhere. Le sigh. Biggest waste of my life. Anyhoo, here it is:
In a tennis club, there are five (5) married couples available to play a "mixed doubles" match, that is, a match in which a combination of one man and one woman play against a combination of another man and another woman. Inhow many ways can a group of four (4) person be chosen for this match if:
i) a man and his wife may play in the match but not as partners?
ii)a man and his wife may not play in the match either as partners of opponents?
The answers are: i) 130 and ii) 60.
I remember doing the question at tutor but when I was flipping back through my book, I had no idea what my working out said. For the question part, the working out was:
5C2 x (4 + 3 x 3) = 130
Dafuq, how does that explain anything?
Okay, that's all. If there's someone out there who will take pity on me, again, I would be really grateful. Thanks :D
the wording of the question is a bit odd :/ are you sure it's meant to be "may play"? i don't think you get the answer of 130 if it's "may play". assuming it's meant to be "must play", the 3rd post of this thread explains the first part of the question:
http://www.sosmath.com/CBB/viewtopic.php?p=89203
for part ii):
lets start with men. there are 5C2 = 10 choices of pairs men. that leaves the women.
lets call the men A, B, C, D, E and the women be a, b, c, d, e
if the pair of men be A & B, that leaves c, d, e as the possible women for the match.
the possible pairings for women would be 3x2=6
thus 10x6=60. so i think that's how you do it?
THANK YOU SO MUCH FIONA!!! That makes more sense now - although if it came up in the actual HSC, I would still have no chance of getting it LOL.