Tennis Club Problem

Mar 2015
4
0
Aberdeen, UK
I'm doing some studying for starting a maths degree in Sept next year and came across this problem in a textbook I'm using for revision.

A tennis club has 16 men and 21 women in it. There are 6 married couples in the club. A team of 6 has to be chosen for a tournament that includes 3 of each sex. How many possible teams of 6 can be chosen if it cannot contain any married couple?

I've tried to get the answer that is provided, but my calculations are off somewhere. The answer is meant to be 629070, but I'm getting 637
790 and I'm not sure where the problem lies. Thanks for any help!
 

Hoempa

Math Team
Apr 2010
2,780
361
Did you assume marriages between a man and a woman? Between men? Between women?
 
Mar 2015
4
0
Aberdeen, UK
Re: Men and women

Yes, it is assumed that only male-female marriages are possible. It's an outdated textbook! ;)
 

CRGreathouse

Forum Staff
Nov 2006
16,046
936
UTC -5
A tennis club has 16 men and 21 women in it. There are 6 married couples in the club. A team of 6 has to be chosen for a tournament that includes 3 of each sex. How many possible teams of 6 can be chosen if it cannot contain any married couple?
If you assume that the marriages are hetero, you have 10 unmarried men and 15 unmarried women, giving
$$
{10\choose3}\cdot{15\choose3} = 54600
$$

If you assume that all the married couples are men, you have 4 unmarried men and 21 unmarried women, giving
$$
{4\choose3}\cdot{21\choose3} = 5320
$$

If you assume that all married couples are women, you have 16 unmarried men and 9 unmarried women, giving
$$
{16\choose3}\cdot{9\choose3} = 47040
$$

The 'most favorable' (largest number of teams) case is two lesbian marriages and four hetero marriages (or 3 and 3), giving
$$
{12\choose3}\cdot{13\choose3} = 62920
$$

If you sum all the possibilities ("we don't know what sex the couples are") you get
$$
\sum_{n=4}^{16} {n\choose3}\cdot{25-n\choose3} = 551980
$$

All of these fall short of the desired 629070.
 
Mar 2015
4
0
Aberdeen, UK
Married people are allowed in the team, it's just that the team becomes invalid if their other half is included as one of the remaining 5 team members. If the married men are {A,B,C,D,E,F} and the married women are {a,b,c,d,e,f}, then the team {A,B,C,d,e,f} is viable because nobody is married to any of their team members.
 

CRGreathouse

Forum Staff
Nov 2006
16,046
936
UTC -5
Married people are allowed in the team, it's just that the team becomes invalid if their other half is included as one of the remaining 5 team members. If the married men are {A,B,C,D,E,F} and the married women are {a,b,c,d,e,f}, then the team {A,B,C,d,e,f} is viable because nobody is married to any of their team members.
Ah, I see.

There will be 0 to 3 married men on the team. Each married man decreases the number of eligible women by 1.
$$
\sum_{n=0}^3 {6\choose n} \cdot {10\choose3-n} \cdot {21-n\choose3} = 629070.
$$
 
  • Like
Reactions: 1 person
Mar 2015
4
0
Aberdeen, UK
Ah, I see.

There will be 0 to 3 married men on the team. Each married man decreases the number of eligible women by 1.
$$
\sum_{n=0}^3 {6\choose n} \cdot {10\choose3-n} \cdot {21-n\choose3} = 629070.
$$
Thanks! I was trying to subtract the number of ways that there were 3, 2 or 1 married couples in the selection from the total number of possibilities, but it was a much more complicated way to do it and I must have a mistake somewhere. This is a much more elegant way to do it. :)