Maxirosario2012
If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?
(1) Combinatorial approach:
\(C^5_3*(C^2_1)^3 =\frac{5*4*3}{3*2}* 2^3 = 10*8 = 80\)
i.e., IF AB, CD, EF, GH, IJ are the couples you have to select only one from a group
1. So selecting one from a group of 2 can be done in \(2C1\)ways. i.e., in 2 ways You have to select 3 people that way. So the total number of possibilities is \(2*2*2 = 8\)
2. Each group of 2 itself has to be selected from 5 such groups. You are selecting 3 groups of 2 from 5 such groups. therefore the total number of possibilities for this is \(5C3= 10\).
3. \((1) * (2) = 8* 10 = 80.\)
Quote:
(2) Reversal combinatorial approach:
Total number of groups: \(C^10_3 = 120\)
Total number of groups with married people: \(C^5_1 * C^4_1 * C^2_1 = 5*4*2 = 40\)
120 - 40 = 80
1. Total number of possibilities of selecting 3 people out of 10 people\(= 10C3= 120\)
2. For total number of groups with a married couple the situation is you have (i)2 people who are married to each other i.e., a group of 2 married to each other and (ii) one other person. The number of possibilities of (i) is the number of ways one group of married people can be selected from 5 groups of married people which is \(5C1 = 5\)ways, the number of possibilities of (ii) can be arrived by finding out in how many ways the remaining person can be selected. It can be done in \(8C1=8\) ways because if you remove the selected married couple 8 persons will remain.
3. Total number of ways of having a married couple\(in the group of 3 = 5*8=40\)4. So number of groups in which 2 people are not married couple \(= 120-40=80.\)