# no. of ways to select $15$ teams from $15$ men and $15$ women such that each team consists of a man and a woman

"Number of ways to select $15$ teams from $15$ men and $15$ women such that each team consists of a man and a woman."

if we keep all the men in a line and make different permutations of women and associate them with the men,,,,,somewhat like having 15 buckets and 15 different fruits to put in them,,,,in this way the answer should be 15! .But the options are 1120,1240,1880,1960

• The answer depends on how you decide if you count different groups of 15 as really different. If you see all 30 people as individuals then it is just the standard question of picking 15 people out of 30 and the genders are a red herring. If on the other hand you say that group A and group B are the same (for the purposes of the question) if both consist of 4 men and 11 women, even if none of the men in group A and only 9 of the women in group A are also in group B then the answer is much much smaller. Mar 23, 2017 at 12:06
• Welcome to MathSE. When you pose a question here, it is expected that you include your own thoughts on the problem. Please edit your question to indicate what you have tried and where you are stuck. Mar 23, 2017 at 12:14
• @N.F.Taussig thanks sir for your advice,,i have done ir Mar 23, 2017 at 12:22
• Your answer to the restated question is correct. There appears to be an error in the answer key. Mar 23, 2017 at 12:53

Total people are 30 and we have to choose 15 by using combinatorics combination formula ans is 30C15 .for eg let there be 15 boxes in each box u want to place a person .Now the first box has 30 option second has 29 third has 28 .....last has 15 so total options =30*29*28...........*15= 30!/15!

• options are 1120,1240,1880,1960 Mar 23, 2017 at 12:09
• sorry ,,i missed a vital part in the question,,i have edited it now. sorry Mar 23, 2017 at 12:10
• Can tell me the question properly Mar 23, 2017 at 12:10
• Can you tell that how many people are there in one team and are all the men different or same Mar 23, 2017 at 12:11
• i have edited the question ,,thats all i can say,,, Mar 23, 2017 at 12:15

The answer is 1240 for the first can be selected as 15C1*15C1 second team can be selected by 14C1*14C1 and so on last team can be selected by 1C1*1C1 total ways = 15C15*15C15+14C14*14c14.....1C1*1C1 = 1240

• according to this logic if we have 2 men and 2 women ,,no. of ways would be 2*2+1*1=5. Sir , can u please tell me those 5 combinations? Mar 23, 2017 at 12:32
• For 2 men and 2 women there would 2 teams first team members can be chosen as 2C1*2C1 =4 now there are 1man and 1 woman and we have to make one more team and we have to take both man and woman in the team this has only 1 way so total ways = 4+1=5 Mar 23, 2017 at 12:36
• if we take the men as A and B and women as 1 and 2,,the the combinations are (A-1 B-2),(A-2 B-1) Mar 23, 2017 at 12:37
• what are the other three combinations ? Mar 23, 2017 at 12:45
• This the gap between practical and theoretical maths Mar 23, 2017 at 12:46