Six couples are invited to play a game of cards, which needs 4 players

Question

Six couples are invited to play a game of cards, which needs 4 players at a time. In how many ways can the players be selected, if no couple should be included?

(A) 256 
(B) 384 
(C) 240 
(D) 320 
(E) 60

IanStewart · Accepted Answer

Imagine first we're picking a first player, second player, third player and fourth player. We'd have 12 choices for the first player, 10 for the second (to avoid picking a couple), 8 for the third and 6 for the fourth. So we'd have (12)(10)(8)(6) choices. But presumably the order of the 4 players is irrelevant, so we must divide by 4!, and we get

(12)(10)(8)(6)/(4!) = (5)(8)(6) = 240 possible selections

(12)(10)(8)(6)/(4!) = (5)(8)(6) = 240 possible selections

RajatJ79 · Answer

Six couples but no couple can play card game at a time. i.e. only a single person form each couple pair
Also, only 4 players can play the game at a time.

Step 1 
We start with selecting 4 couples out of 6 couples. (Because only 4 players can play at one time.)

This is done in  6C4  ways. ===>  6C4 = 6!/(4!*2!)   =  15 ways.

Step 2 
Now that we have selected the 4 couples, we select one from the pair who actually plays the card game.
(Because no couple should be included i.e. only one person from a couple pair can be selected)

4 couples, we can select 1 person from each pair. i.e. 2 choices per couple.
This is done is  2^4  =  2*2*2*2  =  16 ways. 
 
Both conditions need to be satisfied, so it's an AND case of combination.
Total ways = 15*16 =  240 ways .

Answer C.

satya2029 · Answer

Total cases= 12C4=495
Cases which are not required= 2 couples or 1 couple and 2 others(not couple)
2 couples= 6C2=15
1 couple + 2 others=6C1*(1*8C1*5)=240 ( after selecting one couple among 6, we are left with 5 couples. Now select anyone( 1 way), then we are left with 8 people as we can not select the his or her partner. And we can select 1 person among 8 people in 8C1 ways and we have 5 such couples. So 1*8C1*5)
Cases which are not required=255
Required cases=495-255
=240
C:)

Codebug4it · Answer

Answer: Option C

No couples allowed at one time, so one player from a couple can be selected at one time

=> 6C4 =\frac{ 6!}{(4! * 2!)}
=> 15 ways

Each player from a couple can be selected in 2 ways => either first or second and there are 4 couples so,   2 ^ 4 = 16 ways

Total ways =  15 * 16 = 240 ways  (C)

Keyurneema · Answer

* Select 4 couple out of 6 couple- 6C4, & Than selecting 1 person from a couple because both can not Play together- 2 WAYS

=6C4*2*2*2*2= 15*16= 240

Ans -C

Archit3110 · Answer

method 1
12*10*8*6/4! = 240
method 2 
6c3*2*2*2*2 ; 15*16 ; 240
OPTION C

1616 · Answer

Hello  Bunuel ,

Can you please share the solution to this question?

Bunuel · Answer

Six couples are invited to play a game of cards, which needs 4 players at a time. In how many ways can the players be selected, if no couple should be included?

(A) 256
(B) 384
(C) 240
(D) 320
(E) 60

Each couple can send only one "representative" to the game. We can select 4 couples (since there should be 4 players) to send one representative each to the game in  C^4_6  ways.

However, each of these 4 chosen couples can send either the husband or the wife, resulting in  2*2*2*2=2^4  possibilities.

So, the number of ways to choose 4 people from 6 married couples such that none of them are married to each other is:  C^4_6*2^4=240 .

Answer: C.

Feb2024 · Answer

Bunuel   IanStewart   MartyMurray   KarishmaB   chetan2u   gmatophobia

Hello Experts!

Why is the following approach incorrect:

One can choose 3 couples of out 6 in 6C3 ways = 20 ways. 
Next, 3 people of these 3 couples can be choosen in  2*2*2 = 8  ways.

Now, we have 1 vacant position and remaining 6 available to choose from.
So, the other player can be chosen in  6 different ways  to have players from 4 different couples.

Thus number of ways= 20*8*6 = 960

Posted from my mobile device

MartyMurray · Answer

You haven't eliminated all the duplicates.

Notice that the "other" player could have been chosen in the first round of choosing three couples as well. So, each player could be chosen in the couples round or in the final round, and you haven't accounted for that duplication.

bumpbot · Answer

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Six couples are invited to play a game of cards, which needs 4 players

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