Which set of preferences for three candidates is impossible? Hi recently i appeared in an aptitude,there was a problem that i realy cant understand please provide some idea, how to solve it.( sorry to for poor English.)
(Question)-> Three candidates, Amar, Birendra and Chanchal stand for the local election. Opinion polls are conducted and show that fraction a of the voters prefer Amar to Birendra, fraction b prefer Birendra to Chanchal and fraction c prefer Chanchal to Amar. Which of the following is impossible? 
(a) (a, b, c) = (0.51, 0.51, 0.51);
(b) (a, b, c) = (0.61, 0.71, 0.67);
(c) (a, b, c) = (0.68, 0.68, 0.68);
(d) (a, b, c) = (0.49, 0.49, 0.49);
(e) None of the above.

 A: You should consider all six possible orderings (by preference) of the three candidates: $ABC,ACB,BAC,BCA, CAB,CBA$. (Assuming no "ties" occur in the people's opinion).
The question is: Can you assign nonnegative numbers to these six orderings ins such a way that $a=ABC+ACB+CAB$, $b=ABC+BAC+BCA$, and $c=BCA+CAB+CBA$?
For example, we see that $a+b+c=2ABC+ACB+BAC+2BCA+ 2CAB+CBA$, which implies that $1\le a+b+c\le 2$ is a necessary condition.
A: Base on the reasoning of my other answer I was leaded to the following answer that is much simpler.  It show that the condition stated by @HagenvonEitzen is sufficient bei constructiong explicit solutions. 
Lemma: $(a,b,c)$ are possible number (as defined in the OP) 
if and only if $0 \le a+b+c \le 1$.
I assume that people are asked whom they like most and whom they like least of A, B and C and the most liked cannot be the same as the least liked. We have to solve a system of linear equations an inequalities.
We define the three letter variable  as $abc$ the relative frequency that A is liked most and C is liked least by a person. This should mean A is preferred to 
B and C and B to C. 
We now get the folowing four equations:
$$\begin{eqnarray} 
\tag{1}\\
abc + acb + cab & = & a  \\
bac + bca + abc & = & b \\
cba + cab +bca & = & c \\
abc + acb + bca + bac + cab + cba & = & 1
\end{eqnarray}
$$
From this we get
$$a+b+c \\=(abc + acb + cab)+(bac + bca + abc)+(cba + cab +bca)\\=(abc + acb + bca + bac + cab + cba)+(abc+bca+cab)\\=1+(abc+bca+cab)$$
We can conlude that 
$$1 \le a+b+c \le 2 \tag{2}$$
because 
$$0 \le abc+bca+cab \le 1$$
So this is a necessary condition for the problem to have a solution.
Now lets assume that
$(2)$ is valid.
Now let's check the three sums $a+b$, $a+c$, $b+c$.
Without loss of generality we assume that $a \le b \le c$ and  therefore $a+b \le a+c \le b+c$. 
If all of these sums are $ \le 1$ we set
$$acb=1-b,cba=b+c-1,bac=b+a-1,bca=-b-c-a+2,cab=0,abc=0$$
if all of these sums are $ \ge 1$ we set
$$acb=b+c+a-1,cba=0,bac=0,bca=b,cab=-b-a+1,abc=-b-c+1$$
the remaining case is that at least one sum is $\ge 1$ and at least one sum is $\le 1$. 
So we have $a+b \le 1$ and $b+c \ge 1$. We set
$$acb=a,cba=b+c-1,bac=0,bca=1-c,cab=-b-a+1,abc=0$$
It is easy to check that all these defined values lie between $0$ and $1$. So they are valid relative frequencies.
And they satisfy $(1)$. 
So the condition (2) is also sufficient.
In your sample only the answer c does not fullfill the condition.
