according to me 16-6=10 should be answer

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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

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

6 votes

We can think like this,

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

then ,from every couple,select one out of two,i.e ^{2}C_{1 }* ^{2}C_{1 }* ^{2}C_{1 }* ^{2}C_{1 }.

Hence result is ^{4}C_{4} * ^{2}C_{1 }* ^{2}C_{1 }* ^{2}C_{1 }* ^{2}C_{1 }.=16

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???