Problem Solving

A committee of 3 ppl is to be chosen from 4 married couple. What is the no of ways that can be chosen if two people who are married to each other does not serve on the same committee?
A-16 B-24 c-26 D-30 E-32. Can someone give me an explanation for this?

Simplest Way

1)Total Number of ways a committee of 3 can be chosen from 8 ppl with no restrictions=8C3=56 Ways
2)Total Number of ways a committee of 3 can be chosen which will always include one couple=4C1*6C1=24

Statement 1 - Statement 2=56-24=32 Hence E

Alternate Way
Let the ppl be A1 A2 B1 B2 C1 C2 D1 D2

Number of ways to choose the first person is 8 as any one can be chosen
Number of ways to choose the 2nd person (any one besides the spouse of the 1st) i.e. 6 Ways
Number of ways to choose the 3rd person (any one besides the spose of the first two ppl) i.e. 4 ways
However, we have considered ordered arrangements in this method hence, we divide by 3! to get rid of the ordered arrangements. 8*6*4/6= 32 Hence E

Say A-B, C-D, E-F, G-H are the pairs.

No of ways of choosing the first member - 8 (A,B,C,D,E,F,G,H), lets say A is the chosen one

No of ways of choosing the second member - 6 (C,D,E,F,G,H, as you are not allowed to choose B, the partner), lets say C is the chosen one

No of ways of choosing the Third member - 4 (E,F,G,H, as you are not allowed to choose D, the partner,lets say E is the chosen one

So total number = 8*6*4 ways

But as order doesn't matter(ACE is same as AEC, CEA, CAE, EAC, ECA), the total number of ways [spoiler]= 8*6*4/3! = 8*6*4/6 = 32 Answer : Option E[/spoiler]

4 Couples

Committee of 3

__ __ __


Total no of ways of chosing the committee => 8C3 ==> 8!/5!3! ==> 56
No of ways of selecting a couple for the committee of 3 ==> 4C3 * 6 [ 4C3 ways of selecting a cpouple && 6 choices for the 3rd member in each case]
==> 4*6 == 24

No of ways of selecting a committee such taht there are no couples in the committee==> 56 - 24 ==> 32

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