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Tradition tells us that in certain rural areas of Russia the marriage of a young woman was determined as follows: The young woman held in her hand 6 ribbons by the middle, so that the tips were above and below the hand. The young suitor had to tie by pairs the 6 tips that went up, and then tie the 6 tips below, also by pairs. If the young man tied the 6 ribbons in a single circle, then the wedding would take place in less than a year.

a) What is the probability of forming a single circle if the ribbons were tied randomly?

My answer was: (6C2)(4C2)(2C2)/6! However, the answer is: (4C1)(2C1)/[(5C1)(3C1)]

Can someone please explain to me why?

I am using the formula for combinations, nCr = n! / (r! * (n - r)!), where n represents the number of items, and r represents the number of items being chosen at a time

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  • $\begingroup$ What do you mean by 6C2 etc.? $\endgroup$ – pendermath Feb 9 at 17:27
  • $\begingroup$ 6 choose 2 I guess $\endgroup$ – pendermath Feb 9 at 17:28
  • $\begingroup$ nCr (combinations of 2 ribbons in 6 without order) $\endgroup$ – Lollipop Feb 9 at 17:28
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I can confirm that the answer is ${8\over15}$ but I don't understand how the answer with combinations is arrived at. I would do it this way:

After the tops have been tied, we have $3$ ribbons, with $6$ ends. The man picks one end, then ties it to one of the $5$ remaining ends. Four of the five don't belong to the same ribbon, and are okay. If he succeeds in the first step, there are now $2$ ribbons, and by the same argument he has probability of ${2\over3}$ of success. If he succeeds at the the second step there is only one ribbon left, so he must succeed at the third step.

This gives a probability of success of $$\frac45\cdot\frac23={8\over15}$$

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