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Four couples decide to form a committee of  four members. The number of different committees that can be formed in which no couple finds a place is?      

  1. 10
  2. 12
  3. 14
  4. 16
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@Arjun sir wat does it mean by no couple finds a place
according to me 16-6=10 should be answer
–2
–2
Why -6?
0
0

6 Answers

6 votes
6 votes
Best answer
when all four are mens

1. M1 M2 M3 M4

when all four are womens

2. F1 F2 F3 F4

When one man and 3 womens  - C(4,1) = 4

3. M1 F2 F3 F4  

4. M2 F1 F3 F4

5. M3

6. M4

When one woman and 3 mens - C(4,1) = 4

7. F1 M2 M3 M4

8. F2

9. F3

10. F4

When 2 mens and 2 womens, total possibilities= C(4,2)=6.

11. M1 M2 F3 F4

12. M1 M3 F2 F4

13. M1 M4 F2 F3

14. M2 M3 F1 F4

15. M2 M4 F1 F3

16. M3 M4 F1 F2

Ans- 16
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6 votes
6 votes

We can think like this,

Out of the 4 couples,no of ways of selecting 4 couples is 4C4      (since 4 members are required and no two from a single couple)

then ,from every couple,select one out of two,i.e  2C1 * 2C1 * 2C1 * 2C1 .

Hence result is 4C4 * 2C1 * 2C1 * 2C1 * 2C1 .=16

3 votes
3 votes

1):
Say M1 M2 M3 M4 are men and W1 W2 W3 W4 are women.

S1=No. of ways to chose first member is 8.

Now suppose selected member is M1 so W1 would automatically exclude from further selection.

2)

Now we have  M2 M3 M4 W2 W3 W4

S2=No. of ways to chose Second member is 6.

Now suppose selected member is M2 so W2 would automatically exclude from further selection.

Now we left with   M3 M4 W3 W4

same for 3rd memeber

3)

S3= 4

4)

S4=2

Total s1*s2*s3*s4 way if ordering matter

IF ordering doesnot matter than : $\frac{S1*S2*s3*s4}{4!}$

$\frac{8*6*4*2}{4!\frac{}{}}$ ans is 16

Arjun sir i m doubtful this approach is orrect or not???

2 votes
2 votes
To satisfy the question we need to take exactly one person from couple
So among four couples , we can distribute like this

2 - men or women

2*2*2*2
=16
Answer:

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