2,452 views
4 votes
4 votes
For a game in which 2 partners oppose 2 other partners, six men are available. If every possible pair must play against every other pair, the number of games to be played is

 

9a) 36 (b) 45 (c) 42 (d) 90

3 Answers

Best answer
11 votes
11 votes

45 is correct. 

Steps :

1) Choose 4 players who will play in Game = 6C= 15 ways

2) Now, suppose we have 4 players {A,B,C,D} & we have to form two teams of 2.

    means partition of 4 in 2,2

 =>    (AB -   CD) ,  (AC - BD) ,   (AD - BC)   = 3 ways   


Total 15 * 3 = 45 ways (Ans)

selected by
0 votes
0 votes

Among 6 men

2 partner plays game and 2 partner oppose

So,from 6 we can select 2 partner in 6C2 ways =6*5/2=15 ways

Now, there are 4 men

We can select the partner who oppose in 4C2 = 6 ways

So, total number of selection 15*6 = 90

but to remove duplication ans will be  90/2 =45

Answer will be (D) 

edited by

Related questions

2 votes
2 votes
4 answers
1
jaydip74 asked Jul 22, 2023
462 views
In how many ways can 3 non-negative integers be chosen such that a + b + c = 10 where a >= -1 , b >= -5 and c >= 3 ? 3666105None
0 votes
0 votes
2 answers
3
Lakshman Bhaiya asked Oct 30, 2018
1,671 views
9 different books are to be arranged on a bookshelf. 4 of these books were written by Shakespeare, 2 by Dickens, and 3 by Conrad. How many possible permutations are there...
0 votes
0 votes
1 answer
4
eyeamgj asked Oct 23, 2018
799 views
How many ways are there to select five bills from a cash box containing $1 bills, $2 bills, $5bills, $10 bills, $20 bills, $50 bills, and $100 bills? Assume that the orde...