In how many ways can a mixed doubles badminton game can be organized

Question

In how many ways can a mixed doubles badminton game be organized among 6 men – A, B, C, D, E, and F and 4 women – P, Q, R, and S such that A and P cannot play in the same game?

(Note: A game happens between 2 teams and in a mixed-doubles game a man and a woman pair up to play against another team of one man and one woman)

A. 30

B. 90

C. 120

D. 150

E. 180

ChandlerBong · Answer

Hi Experts Bunuel , ThatDudeKnows , KarishmaB , ScottTargetTestPrep , Can you please share your approach to this combination problem? Thanks in advance.

Bhupendra46 · Answer

The total number of ways we can organize a match is 6C2 * 4C2 = 90
We can form 2 different combination of mixed team from 2 men and 2 women. So the total number would be 2*90 = 180.

We want that A and P should note be in a match togather, so we will find number of ways in which A and P are togather and subtract it from total 180.

So now we have fixed that A and P are in a match so , now we only have to select other two members.
5C1 * 3C1 = 15.
Same as previous two combinations 15*2 = 30.
Final answer 180 - 30 = 150
Options D.

Posted from my mobile device

Bunuel · Answer

Men: A, B, C, D, E, F
Women: P, Q, R, S

First, let's calculate the total number of games possible without any restrictions.

Choosing two men from six (6C2) gives us 15 combinations. These represent the two male players in two opposing teams. For instance, (A, _) vs (B, _).

Choosing two women from four (4C2) gives us 6 combinations. These represent the two female players who can be paired with the selected two men. However, there are two ways these women can be matched with the men, hence we multiply by 2. For instance, P and Q with A and B can be paired either as (A, P) vs (B, Q) or as (A, Q) vs (B, P).

Therefore, the total number of unrestricted games would be 15 (male pairs) * 6 (female pairs) * 2 (arrangements) = 180.

Now, let's consider the restriction that A and P cannot play in the same game.

If A and P are on the same team (A, P), the number of opposing pairs will be 5*3 = 15. This is because any of the remaining 5 men can be paired with any of the remaining 3 women. For instance, (A, P) vs (B, Q); (A, P) vs (B, R); (A, P) vs (B, S); etc.

If A and P are on opposing teams (A, _) vs (_, P), the number of opposing pairs will also be 3*5 = 15. This is because A can be paired with any of the remaining 3 women and P can be paired with any of the remaining 5 men. For instance, (A, Q) vs (B, P); (A, Q) vs (C, P), etc.

Therefore, the total number of restricted games equals the total number of unrestricted games minus the restricted games: 180 - (15 + 15) = 150.

Answer: D.

gmatophobia · Answer

Constraint Given  : A and P cannot play in the same game

To form two teams we need 2 men and 2 women.

1) Select two men from 6 men =  ^6C_2  = 15
1) Select two women from 4 women =  ^4C_2  = 6

Total number of games = 15 * 6 * 2 = 180

However, this also includes the possibility when A and P are a part of the team. Let's subtract those cases -

Cases in which A and P are a part of the game

If A is already a part of the team we need to select 1 man out of 5 available men. This can be done in  ^5C_1  ways
If P is already a part of the team we need to select 1 woman out of 3 available women. This can be done in  ^3C_1  ways

Number of ways of forming a team such that A and P are part of the same game = 5 * 3 * 2 = 30

Number of ways to form a team such that A and P cannot play in the same game = 180 - 30 = 150

Option D

thelastskybender · Answer

No. of games in which A and P do not play in the same game= 
 All - No. of games ( A and P both play together) 
= (6C2 x 4C2)x 2 - (5C1 x 3C1)x2  x2 because a man can play with either of the two women selected]  
= 150

KarishmaB · Answer

Here is what I would do:

No of ways of selecting team 1 = 6 * 4
No of ways of selecting team 2 = 5 * 3 
But there are no "team 1 and team 2" i.e. we need to un-arrange so divide by 2

No of ways to make 2 teams to play without constraints = 6 * 4 * 5 * 3/2 = 180

A and P should not play together. Select another man and a woman in 5 * 3 ways (say B and R) and multiply by 2 for the two ways in which you can split them into 2 teams - {AP and BR} or {AR and BP}
This gives us 30 ways.

Acceptable number of ways = 180 - 30 = 150

Answer (D)

* When they say "do not put them together," put them together and subtract out those ways.

bumpbot · Answer

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In how many ways can a mixed doubles badminton game can be organized

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