1. ## Combination question about forming mixed hockey

Question: In a game of mixed hockey there are ten married couples and 2 spinsters playing. In how many ways can the two teams be made up, if no husband may play against his wife?

case 1:
team 1 (include 2 spinsters)

2 spinsters ; 2C2 =1
there are 4 spaces to fill, we need 2 couples to fill it; 10C2 =45
2C2 x 10 C2 = 45

team 2;
there are 6 spaces ( 3 couples to fill it)
8C3 = 56

case 2:
team 1 ( no spinsters)
we need 3 couples to fill it;
10C3 = 120

team 2
there are 2 options include 2 spinsters or not

if we include 2 spinsters; 2C2 =1
and we are left with 4 spaces to fill it; 7C2 = 21

if we do not include spinsters;

then we need 3 couples; 7C3 = 35

sum ; 45+120+56+35+21 = 277

I do not know where I went wrong?

2. ## Re: Combination question about forming mixed hockey

Originally Posted by bigmansouf
Question: In a game of mixed hockey there are ten married couples and 2 spinsters playing. In how many ways can the two teams be made up, if no husband may play against his wife?
I am clueless as to anything about hockey. Please tell us how one forms a hockey team? How many on a team?
Moreover, can a husband & wife be on the same team?

3. ## Re: Combination question about forming mixed hockey

Ice hockey has 6 people per team, Field hockey has 11 people per team.

Given the wording of the question, i.e. "the two teams", I think they are referring to field hockey. This would use all the people mentioned.

A team then must be comprised of 5 couples and 1 spinster. Otherwise some couple would be playing against each other.

given this the answer should be simply

$N = \dbinom{10}{5}\cdot \dbinom{2}{1} = 504$

This is precisely double the answer given. I can't account for the discrepancy. I think it quite possible the answer given is wrong.

4. ## Re: Combination question about forming mixed hockey

Originally Posted by Plato
I am clueless as to anything about hockey. Please tell us how one forms a hockey team? How many on a team?
Moreover, can a husband & wife be on the same team?
i thought that it was about ice hockey but it is about field hockey

5. ## Re: Combination question about forming mixed hockey

Originally Posted by romsek
Ice hockey has 6 people per team, Field hockey has 11 people per team.

Given the wording of the question, i.e. "the two teams", I think they are referring to field hockey. This would use all the people mentioned.

A team then must be comprised of 5 couples and 1 spinster. Otherwise some couple would be playing against each other.

given this the answer should be simply

$N = \dbinom{10}{5}\cdot \dbinom{2}{1} = 504$

This is precisely double the answer given. I can't account for the discrepancy. I think it quite possible the answer given is wrong.
thank you very much i am having problems with this kind of questions
such as; "How many mixed hockey teams may be made from 6 married couples, 1 bachelor and 3 spinsters, if no wife will play without her husband? "
for this question;
i realise that there are two ways to do this 4 couples and 3 individuals and 5 couples and 1 individual

4 couples and 3 individuals;
6C4 = 15 4C3 = 4 15 x 4 = 60

5 couples and 1 individual
6C5 =6 4C1= 4 6 x 4 = 24

the answer is 60 + 24 = 84 but the book gives an answer of 876 but i keep getting the answer wrong

please can you help me and how do i improve for these kind of questions

6. ## Re: Combination question about forming mixed hockey

Originally Posted by bigmansouf
thank you very much i am having problems with this kind of questions
such as; "How many mixed hockey teams may be made from 6 married couples, 1 bachelor and 3 spinsters, if no wife will play without her husband? "
for this question;
i realise that there are two ways to do this 4 couples and 3 individuals and 5 couples and 1 individual

4 couples and 3 individuals;
6C4 = 15 4C3 = 4 15 x 4 = 60

5 couples and 1 individual
6C5 =6 4C1= 4 6 x 4 = 24

the answer is 60 + 24 = 84 but the book gives an answer of 876 but i keep getting the answer wrong

please can you help me and how do i improve for these kind of questions
One thing you've neglected is that single husbands can play on the team w/o their wives.

$\left( \begin{array}{ccc} H& C & S \\ 0 & 5 & 1 \\ 1 & 5 & 0 \\ 0 & 4 & 3 \\ 1 & 4 & 2 \\ 2 & 4 & 1 \\ 1 & 3 & 4 \\ 2 & 3 & 3 \\ 3 & 3 & 2 \\ 4 & 2 & 3 \\ 3 & 2 & 4 \\ 5 & 1 & 4 \\ \end{array} \right)$

For each triple compute $\dbinom{6}{H}\dbinom{6-H}{C}\dbinom{4}{S}$ and sum them all up.
This sums to 876.

7. ## Re: Combination question about forming mixed hockey

Originally Posted by romsek
Ice hockey has 6 people per team, Field hockey has 11 people per team.

Given the wording of the question, i.e. "the two teams", I think they are referring to field hockey. This would use all the people mentioned.

A team then must be comprised of 5 couples and 1 spinster. Otherwise some couple would be playing against each other.

given this the answer should be simply

$N = \dbinom{10}{5}\cdot \dbinom{2}{1} = 504$

This is precisely double the answer given. I can't account for the discrepancy. I think it quite possible the answer given is wrong.
Thinking about this I guess the factor of two comes from the fact that they don't care about the order the two teams are formed in.

8. ## Re: Combination question about forming mixed hockey

Originally Posted by romsek
One thing you've neglected is that single husbands can play on the team w/o their wives.

$\left( \begin{array}{ccc} H& C & S \\ 0 & 5 & 1 \\ 1 & 5 & 0 \\ 0 & 4 & 3 \\ 1 & 4 & 2 \\ 2 & 4 & 1 \\ 1 & 3 & 4 \\ 2 & 3 & 3 \\ 3 & 3 & 2 \\ 4 & 2 & 3 \\ 3 & 2 & 4 \\ 5 & 1 & 4 \\ \end{array} \right)$

For each triple compute $\dbinom{6}{H}\dbinom{6-H}{C}\dbinom{4}{S}$ and sum them all up.
This sums to 876.
thank you very much

for to be a bother but can you explain what C stands for in the table
H- Husband
C- ?
S- spinsters

is C for Bachelors

9. ## Re: Combination question about forming mixed hockey

Originally Posted by bigmansouf
thank you very much

for to be a bother but can you explain what C stands for in the table
H- Husband
C- ?
S- spinsters

is C for Bachelors
H is a husband taken by himself
C is an entire couple
S is for anyone not married (the spinsters and the bachelor)