Probability (1 Viewer)

[Just.Snaz]

In a tennis club, there are five 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. In how many ways can a group of four persons be chosen for this match if:

i. a man and his wife play in a match but not as partners;

ii. a man and his wife may not play in the match either as partners or as opponents.


*shudders*

help anyone? thanks in advance.

 

[dolbinau]

Is the first one 6C4?

I don't know *hangs self*

 

[Just.Snaz]

Is the first one 6C4?

I don't know *hangs self*

nope.. *joins in*

 

[dolbinau]

Guh. Where is this question from?

 

[Just.Snaz]

Catholic trial 1988

curse those trials.

 

[dwarven]

In a tennis club, there are five 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. In how many ways can a group of four persons be chosen for this match if:

i. a man and his wife play in a match but not as partners;

ii. a man and his wife may not play in the match either as partners or as opponents.


*shudders*

help anyone? thanks in advance.

i) 10C4/10C2 :S

 

[Just.Snaz]

i) 10C4/10C2 :S

lol that gets a fraction.. and not close to the answer either..

i'll try call the hsc dudes when the lines open

In the mean time.. anyone who gets it will get a cookie

 

[Azreil]

Don't bother with this question. My teacher, who's the best mathematician I've ever met, couldn't interpret it correctly, and told our class not to bother with it.

Moral of the story: Terrible question.

 

[Just.Snaz]

5C2x5C2 divided by something?

I'm making the assumption that they choose 2 people out of the men and 2 people out of the women instead of 4 people out of the 10 :S

well that's what i was doing.. and then divide by 2 since Ma + Wb playing with Mb + Wa is doubled cause it takes into consideration the side of the courts... i think...

 

[independantz]

In a tennis club, there are five 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. In how many ways can a group of four persons be chosen for this match if:

i. a man and his wife play in a match but not as partners;

ii. a man and his wife may not play in the match either as partners or as opponents.


*shudders*

help anyone? thanks in advance.

i) 8C1.2C1.7C1.1C1
ii)8C4

i think.... If i interpreted the question correctly. Does it mean that a particular man and his wife or all the couples ? I assume it means the first as it says" a man"

 

[Just.Snaz]

i) ((1C1x8C1x1C1x7C1)/2)/10C4
ii)8C4/10C4

i think.... If i interpreted the question correctly. Does it mean that a particular man and his wife or all the couples ? I assume it means the first as it says" a man"

are you working out the probability? cause your answers are fractions.. and it asks for arrangments

what i understand from the question is it's asking how many ways can you make a pair of two couples from 5 married couples. but the couples can't be married.

 

[Just.Snaz]

okay i worked out that for each of the 5 men, there are 4 ways they can be paired with a woman that is not their wife. now for the other spot for a man, there are 4 possibilties, and there are 4 ways each of the 4 men can be paired with a woman that is not his wife.

can anyone convert this to maths?

 

[independantz]

are you working out the probability? cause your answers are fractions.. and it asks for arrangments

Whoops, i jsut read the thread title and got the important information lol.

 

[Just.Snaz]

Whoops, i jsut read the thread title and got the important information lol.

lol my mistake.. I tend to class everything of this sort under that hideous topic..

 

[samwell]

In a tennis club, there are five 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. In how many ways can a group of four persons be chosen for this match if:

i. a man and his wife play in a match but not as partners;

ii. a man and his wife may not play in the match either as partners or as opponents.


*shudders*

help anyone? thanks in advance.

(i)there are 5 men/ for a doubles game there is 5C2 ways of choosing 2 men or women. Then after this there is only one way to choose their couple.
but then they are not to play on the same side.
u can think of this as an arrangement where by if the man is first then his wife is not to come second therefore the expresion goes (2) (1) (1) (1) since its only the first pick u have two options then the others will fall into place.
=my answer is 2*5C2

(ii) the no of ways of choosing either 2men or 2 women is 5C2.
After choosing 1 man u cannot choose his wife. Therefore u now have 3 women left to choose from.
my answer is : 5C2*3C2


Not sure but seems logical

 

[dwarven]

can i have a cookie for trying?