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?

What we'll do is, we'll let Kamrul tie the first three knots, and then label the ropes, and then let him tie the other three.  Thus no matter what ropes he chooses the first time, we can say he has tied A to B, C to D, and E to F.
The number of perfect matchings for $2n$ objects is $(2n-1)!!$, the odd double factorials.  For six objects, this is $5!!=15$.  This is small enough that I can do it out by hand:
AB CD EF: three loops, AB CD EF
AB CE DF: two loops, AB CDFE
AB CF DE: two loops, AB CDEF
AC BD EF: two loops, ABDC EF
AC BE DF: one loop, ABEFDC
AC BF DE: one loop, ABFEDC
AD BC EF: two loops, ABCD EF
AD BE CF: one loop, ABEFCD
AD BF CE: one loop, ABFECD
AE BC DF: one loop, ABCDFE
AE BD CF: one loop, ABDCFE
AE BF CD: two loops, ABFE CD
AF BC DE: one loop, ABCDEF
AF BD CE: one loop, ABDCEF
AF BE CDL two loops, ABEF CD

So (assuming you are counting the ways that all of them will be in a single loop), there are 15 ways of tying the ropes, and 8 of those ways will create a single loop.
