How many ways to tie $2$ ropes so that we do not have a loop BdMO 2014 Higher Secondary:

Avik is holding six identical ropes in his hand where the mid portion of the rope is in his
  fist. The first end of the ropes is lying in one side, and the other ends of the rope are lying
  on another side. Kamrul randomly chooses the end points of the rope from one side, then tie
  every two of them together. And then he did the same thing for the other end. If the
  probability of creating a loop after tying all six rope is expressed as $\dfrac{a}b$, (where a, b are coprime) find the value of $(a+b)$.

There are $\dbinom{6}{2}$ possible pairs of ropes which we could tie.Also,the number of different ways to tie three pairs of ropes is $\dbinom{6}2\cdot\dbinom{4}2$ which is $90$.Hence the total number of ways to tie the ends on both sides(we will call these sides top and bottom) is $90\cdot 90$. Now,consider first tying the ends at the bottom.There are $\dbinom{6}{2}-3=12$ possible ways to tie the ropes at the top.Hence the total number of ways of not having any loop is $\dbinom{12}3\cdot90$.Hence,the probability of not having any loop is $\dfrac{2}{15}$.Thus the required probability is $\dfrac{13}{15}$ and the required sum is $28$. Am I right?

After doing what he did he has three ropes: $1,2,3$. Suppose he puts them on the table, He begins tying rope $1$, the probability he ties it with another rope is $\frac{4}{5}$ since there are $5$ other rope ends and only $1$ is the same rope. If it ties with another rope there are now two ropes, proceed to tie the rope that is not made up of two ropes, the probability it ties to the other rope is $\frac{2}{3}$ After this there are only two edges of one big rope and there is only one way to tie them. The probability is therefore $\frac{4}{5}\cdot\frac{2}{3}=\frac{8}{15}$
