In how many ways can a team of 4 individuals be selected out of 6 coup

Question

In how many ways can a team of 4 individuals be selected out of 6 couples so that no couple is selected

A) 16

B) 30

C) 60

D) 120

E) 240

chetan2u · Accepted Answer

By permutation..

Let the first person be selected in 12C1 or 12 ways.
Next can be any except partner of first person chosen=10C1.. similarly 3rd and 4th in 8C1 and 6C1 ways.
so total = 12*10*8*6
But we are looking for a team where ORDER does not matter, so divide by 4!= \frac{12*10*8*6}{4!}=10*8*3=240

E

Abhi13545001 · Answer

6C4: selecting any 4 couples from 6.
2C1: selecting 1 person from each couple.

here we first selected 4 couples and then one one person from each selected 4.
so, 6C4*2c1*2c1*2c1*2c1= 15*16=240

AntrikshR · Answer

6 couples=12 persons
A1A2,B1B2,C1C2,D1D2,E1E2 and F1F2.
Ways in which 4 persons can be selected out of 12 is 12C4.
Now,Assume you have fixed a couple A1A2,So rest 2 places can be selected in 10C2 ways and similar to A1A2 we can fix any one of the 6 couples. so, ways of selecting 4 persons out 12 such that no couple is selected is 12C4-6*10C2.Now,
Here we have subtracted all the selections where two couples were selected twice so we have to add one back.
(Ex. A1A2B1B2 when we fixed A1A2 and B1B2A1A2 when we fixed B1B2. Both the groups are essentially the same.)
2 couples can be selected in 6C2 ways.
So final answer is 12C4-6*10C2+6C2 =240

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PyjamaScientist · Answer

chetan2u   IanStewart  If the question went like this- There are 6 couples in a room from which 4 people are selected. There are how many ways that there is exactly one couple selected? How would you have done this using your method?

IanStewart · Answer

You would then have 6 choices for the couple. Once you've chosen the couple, you have 10 choices for the third person on the team, and 8 choices for the fourth person, but the order of those two people doesn't matter, so we'd divide by 2!. So we'd have 6 * (10*8/2!) = 240 options in total.

You can confirm that answer in a different way. From the solutions earlier in the thread, we see there are (coincidentally) also 240 ways to choose no couples. If we wanted to choose two couples, we'd have 6 choices for the first, and 5 for the second, but since their order wouldn't matter we'd have 6*5/2! = 15 ways to choose two couples. Since we can't possibly choose more than two couples if we're only picking 4 people, adding the number of ways to choose zero, one and two couples, we find there should be 240 + 240 + 15 = 495 options altogether if we had no restrictions at all on how many couples we choose. But if we have no restrictions at all, we're just picking 4 people from 12, so we have 12C4 = (12)(11)(10)(9)/4! = 11*5*9 = 495 options. So that confirms the answer is correct, and also offers an alternative (but longer) way to solve the problem; to answer your question, you could count all of the possibilities with no restrictions, then subtract the options you don't want to count (zero couples and two couples).

PyjamaScientist · Answer

Thank you  IanStewart  for your support. Really appreciate.

Stanindaw · Answer

Stanindaw · Answer

can somebody please confirm this method coz i am getting 210 as answer by this method

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Regor60 · Answer

6C4 is 6!/4!2! = 6x5/2 = 15

2C1 = 2. 2x2x2x2 = 16

If you're using a different method, show your work and the error can be explained

nonu14 · Answer

I understood this solution, however I solved this ques through following way, Please help me understand where am I going wrong?

no couple= Total-( Exact 1 couple+exact 2 couple)
 = 12C4-(6C1*10C2 + 6C2) = 210

AbhishekP220108 · Answer

Step 1- 4 couples out of 6 in 6c4 ways

Step 2- selecting 1 out of 2 from 1 couple in 2c1 ways. Since there are 4 couples hence (2c1)^4

Step 3- total no of ways= 6c4×(2c1)^4 = 15×16=240

In how many ways can a team of 4 individuals be selected out of 6 coup

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GMAT Problem Solving (PS) Questions

In how many ways can a team of 4 individuals be selected out of 6 coup

Prep Toolkit

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