harithakishore wrote:
But these 4 chosen couples can send two persons (either husband or wife): 2*2*2*2
Bunuel,can you please..please explain this..im confused....
4 chosen couples...i think we can choose 4 different people and not couples..im really confused and also how come it is 2*2*2*2...please explain
We have 6 couples:
\(A (a_1, a_2)\);
\(B (b_1, b_2)\);
\(C (c_1, c_2)\);
\(D (d_1, d_2)\);
\(E (e_1, e_2)\);
\(F (f_1, f_2)\);
We should choose 4 people so that none of them will be married to each other.
The above means that 4 chosen people will be from 4 different couples, for example from A, B, C, D or from A, D, E, F...
The # of ways to choose from which 4 couples these 4 people will be is \(C^4_6=15\);
Let's consider one particular group of 4 couples: {A, B, C, D}. Now, from couple A in the group could be either \(a_1\) or \(a_2\), from couple B in the group could be either \(b_1\) or \(b_2\), from couple C in the group could be either \(c_1\) or \(c_2\), and from couple D in the group could be either \(d_1\) or \(d_2\). So each couple has two options (each couple can be represented in the group of 4 people by \(x_1\) or \(x_2\)), so one particular group of 4 couples {A, B, C, D} can give us \(2*2*2*2=2^4\) groups of 4 people from different couples.
One particular group of 4 couples {A, B, C, D} gives \(2^4\) groups of 4 people from different couples --> 15 groups give \(15*2^4\) groups of 4 people from different couples (total # of ways to choose 4 people so that no two will be from the same couple) .
Hope it's clear.
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