September 10th, 2011, 04:37 AM  #1 
Newbie Joined: Sep 2011 Posts: 7 Thanks: 0  Double 6 Down
Hi I have to confess to having failed to solve this problem that has puzzled me regarding dominoes. I was playing a game with a double 6 set of dominoes called Fives and Threes with 3 friends. The essence of the game is that any ends that can be divided by 3 or/and 5 give points e.g. double six with a 63 attached adds to 15 so = 8 points as it is divided by both 5 and 3 but that information is for interest only at this stage, but what I would like to know is how to calculate the number of possible permutations of the first 4 dominoes can be played following the first player placing the double 6 as the starting domino. e.g one would be 66 then 65 then 53 then 33 how many others are there? 
September 10th, 2011, 06:20 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,629 Thanks: 2077 
What possibilities are you excluding by using the word "starting"? Does the set use blanks as well as digits? Once you've solved the corresponding problem for three dominoes, consider the number of possibilities for the fourth domino, which depends on which have already been used.

September 10th, 2011, 07:00 AM  #3 
Newbie Joined: Sep 2011 Posts: 7 Thanks: 0  Re: Double 6 Down
Hi There are blanks in the set. I am afraid my maths is not really up to solving past the first two dominoes let alone the first 3 and was hoping there was a simple way to calculate this problem and that someone would be kind enough to show the way. My thoughts go only this far and exclude the possibilities of players not being able to play at all. From the first dominoe 66 there are only 6 dominoes that can be played 65,64,63,62,61,60 From here there are (I think as you can play on either end) 11 for each of the above possibilities e.g. for 6665 there would be 54,53,52,51,50,64,63,62,61,60,55 From here I get confused I am now in my 60's and I am afraid my education was very poor 
September 10th, 2011, 07:12 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,629 Thanks: 2077 
Although there are blanks, you're not using any in your examples. The count will depend on the precise rules that apply for how dominoes may be placed. For example, can the same domino be added in slightly different positions? It shouldn't be difficult to list all possibilities for just three dominoes once the rules are known.

September 10th, 2011, 12:25 PM  #5  
Newbie Joined: Sep 2011 Posts: 7 Thanks: 0  Re: Quote:
 
September 11th, 2011, 12:51 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 20,629 Thanks: 2077 
You're right, and your counts so far are correct. Let's start by listing all possibilities for a chain that has the double six at its left end. There are 6 possibilities for the second domino. For each of those possibilities, the third can be a double (leaving 5 possibilities for the fourth), else there are another 5 possibilities for the third domino, for each of which there are 6 possibilities for the fourth. That gives a total of 6(1(5) + 5(6)) = 210. Similarly, there are 210 possibilities for a chain that has the double six at its right end. There remain to be considered the 6(5) = 30 ways of having a domino placed on each side of the double six. For each of these the fourth domino can be to the left or right of the other three and there is a choice of six dominoes for each of those two possibilities, making 30(2)(6) = 360 possibilities in total in addition to the 420 previously counted, so the grand total is 780. There are additional arrangement if, say, you are allowed to have a domino above or below the double six as well as left or right. 
September 11th, 2011, 01:50 AM  #7 
Newbie Joined: Sep 2011 Posts: 7 Thanks: 0  Re: Double 6 Down
Many thanks for that most helpful. As I am only interested in playing two ends, I would now like to know how many possibilities exist in the first round of play where each of the 2nd 3rd and 4 player cannot play as they do not have an appropriate domino! There could be various permutations of this even also.......my head is beginning to hurt. 
September 11th, 2011, 01:56 AM  #8  
Newbie Joined: Sep 2011 Posts: 7 Thanks: 0  Re: Quote:
many thanks  
September 11th, 2011, 04:45 AM  #9 
Global Moderator Joined: Dec 2006 Posts: 20,629 Thanks: 2077 
It's essentially just counting the leaves on a tree, given the branching structure. For a given 2nd domino, there is one possibility (that the 3rd domino is a double) for which there are five possibilities for the fourth domino, and there are five other possibilities for the 3rd domino (you've already given an example), for each of which there are six possibilities for the fourth domino. That's why I added together one five and five sixes. The initial factor of 6 is because there are six possibilities for the second domino. All that counting assumes that the double six is at the extreme left. By symmetry, the count will be the same if the double six is at the extreme right.

September 11th, 2011, 11:23 PM  #10  
Newbie Joined: Sep 2011 Posts: 7 Thanks: 0  Re: Quote:
 

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