Quandaries and Queries Name: trudy Question: A boy holds 6 pieces of strings in his hands with the ends protruding above and below. The top ends are tied together in pairs and then the lower ends are tied together in pairs. What is the probabilty that the pieces of string are all joined in one loop? What is the probability of obtaining two loops? Can you generalise this to solve the problem for 2n blades of grass? Hi Trudy, Look at the six string ends above his hand. Choose any one of them. There are 5 choices for the string to tie it to. Once you have done that there are 4 untied strings. Choose any one of them. There are 3 choices of the string to tie it to. Once that is done you must tie the remaining two strings. Thus there are 531 ways to tie the string ends above his hand in pairs. In a similar way there are 531 ways to tie the string ends below his hand in pairs. Hence there are (531)2 ways to tie the ends together. How many of these result in the strings forming one loop? Take the top end on any string (call it string 1) and choose one of the available five ends to tie to it. There are 5 choices. Call the string you choose, string 2. Now look at the bottom end of string 2. It can't be tied to itself and it can't be tied to the bottom end of string 1 or you will have a loop. Thus there are 4 choices of the string to tie to it. Call the string you choose, string 3. Now look at the top end of string 3. There are three availabe top ends of strings to tie to it, hence 3 choices. Choose one of them and call it string 4. Again, look at the bottom end of string 4. You can't tie it to string 1 or you will complete a loop. Strings 2 and 3 are already tied togethet so there are only two available string ends to tie to it. Choose one of them and call it string 5. Now you are out of choices. The top end of string 5 can only be tied to the one remaining end and the bottom end of that string must be tied to string 1, forming one loop. Hence there are 54321 ways to form one loop. Hence the probability that the pieces of string are all joined in one loop is  54321/(531)2 54321 is called 5 factorial and written 5!. Similarily n!=n(n-1)(n-2)...21. 531 ic called the double factorial of 5 and written 5!!. If n is odd then n!!=n(n-2)(n-4)...1 and if n is even then n!!=n(n-2)(n-4)...2. Using this notation the argument above for 6 strings can be extended to 2n strings to show that the probability that the pieces of string are all joined in one loop is (2n-1)!/[(2n-1)!!]2 Andrei and Penny Go to Math Central