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The function combin(22,6) produces 74,613 combinations.
However, I have an excel program which eleminates the majority of the combination because it allows not more than 2 numbers in each subset to be repeated. This results in a total of 77 combinations that meet the criteria. Example: 1,2,3,4,5,6 1,2,7,8,9,10 1,2,11,12,13,14 1,2,15,16,17,18 1,2,19,20,21,22 1,3,7,11,15,19 1,3,8,12,16,20 How can the combin function be manipulated to show the result of 77? For testing purposes =combin(23,7) produces 245,157 combinations but when not more than 3 numbers are allowed to repeat the result is 253 combinations that meet the criteria. Dk |
#2
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On Oct 31, 12:38*am, Daka wrote:
The function combin(22,6) produces 74,613 combinations. However, I have an excel program which eleminates the majority of the combination because it allows not more than 2 numbers in each subset to be repeated. This results in a total of 77 combinations that meet the criteria. Example: 1,2,3,4,5,6 1,2,7,8,9,10 1,2,11,12,13,14 1,2,15,16,17,18 1,2,19,20,21,22 1,3,7,11,15,19 1,3,8,12,16,20 How can the combin function be manipulated to show the result of 77? For testing purposes =combin(23,7) produces 245,157 combinations but when not more than 3 numbers are allowed to repeat the result is 253 combinations that meet the criteria. Dk Hi, For your first question with 22 and 6 (no more than 2 #'s), the answer is =COMBIN(22,3)/COMBIN(6,3). For your second question with 23 and 7 (no more than 3 #'s), the answer is =COMBIN(23,4)/COMBIN(7,4). In general, for N total numbers of which you select S and allow no more than D numbers to repeat, the answer would be =COMBIN(N,D+1)/ COMBIN(S,D+1). Using the N=23, S=7, D=3 example: As D = 3, every set of 4 elements can only appear 1 time, otherwise it will violate your constraints. The first thing to determine is how many distinct sets of 4 (D+1) can be generated from the 23 (N) numbers. For this example, that is COMBIN(N,D+1) = COMBIN(23,3+1) = 8,855 sets. The second thing to determine is how many distinct sets of 4 (D+1) get used up every time you choose 7 (S) elements. For this example, that is COMBIN(S,D+1) = COMBIN(7,3+1) = 35. Therefore you can have 8,855 / 35 = 253 combinations. S |
#3
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On Nov 3, 11:47*am, Scott wrote:
On Oct 31, 12:38*am, Daka wrote: The function combin(22,6) produces 74,613 combinations. However, I have an excel program which eleminates the majority of the combination because it allows not more than 2 numbers in each subset to be repeated. This results in a total of 77 combinations that meet the criteria. Example: 1,2,3,4,5,6 1,2,7,8,9,10 1,2,11,12,13,14 1,2,15,16,17,18 1,2,19,20,21,22 1,3,7,11,15,19 1,3,8,12,16,20 How can the combin function be manipulated to show the result of 77? For testing purposes =combin(23,7) produces 245,157 combinations but when not more than 3 numbers are allowed to repeat the result is 253 combinations that meet the criteria. Dk Hi, For your first question with 22 and 6 (no more than 2 #'s), the answer is =COMBIN(22,3)/COMBIN(6,3). For your second question with 23 and 7 (no more than 3 #'s), the answer is =COMBIN(23,4)/COMBIN(7,4). In general, for N total numbers of which you select S and allow no more than D numbers to repeat, the answer would be =COMBIN(N,D+1)/ COMBIN(S,D+1). Using the N=23, S=7, D=3 example: As D = 3, every set of 4 elements can only appear 1 time, otherwise it will violate your constraints. The first thing to determine is how many distinct sets of 4 (D+1) can be generated from the 23 (N) numbers. *For this example, that is COMBIN(N,D+1) = COMBIN(23,3+1) = 8,855 sets. *The second thing to determine is how many distinct sets of 4 (D+1) get used up every time you choose 7 (S) elements. *For this example, that is COMBIN(S,D+1) = COMBIN(7,3+1) = 35. *Therefore you can have 8,855 / 35 = 253 combinations. S Oops, sorry, I forgot to mention that this is a maximum. The actual number of combinations the parameters permit when you attempt to construct the set may be lower. I couldn't tell you off the top of my head how many instances or under what conditions it is less than the theoretical optimal number. An example of where it is less is using N=5, S=4, D=2, COMBIN(5,3)/ COMBIN(4,3) = 2.5. When you attempt to construct the combinations, you'll find there is only 1. (Any one of {1,2,3,4}, {1,2,3,5}, {1,2,4,5}, {1,3,4,5}, {1,2,3,4}, but you can't choose 2 of these as you'll end up violating the constraints) [In general, for N = k, S= k-1, and D=2, the actual number of sets is going to be 1.] You'll probably need someone with a bit more insight to help you determine when you will be unable to achieve the theoretical maximum. S |
#4
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On Nov 3, 2:22*pm, Scott wrote:
On Nov 3, 11:47*am, Scott wrote: On Oct 31, 12:38*am, Daka wrote: The function combin(22,6) produces 74,613 combinations. However, I have an excel program which eleminates the majority of the combination because it allows not more than 2 numbers in each subset to be repeated. This results in a total of 77 combinations that meet the criteria. Example: 1,2,3,4,5,6 1,2,7,8,9,10 1,2,11,12,13,14 1,2,15,16,17,18 1,2,19,20,21,22 1,3,7,11,15,19 1,3,8,12,16,20 How can the combin function be manipulated to show the result of 77? For testing purposes =combin(23,7) produces 245,157 combinations but when not more than 3 numbers are allowed to repeat the result is 253 combinations that meet the criteria. Dk Hi, For your first question with 22 and 6 (no more than 2 #'s), the answer is =COMBIN(22,3)/COMBIN(6,3). For your second question with 23 and 7 (no more than 3 #'s), the answer is =COMBIN(23,4)/COMBIN(7,4). In general, for N total numbers of which you select S and allow no more than D numbers to repeat, the answer would be =COMBIN(N,D+1)/ COMBIN(S,D+1). Using the N=23, S=7, D=3 example: As D = 3, every set of 4 elements can only appear 1 time, otherwise it will violate your constraints. The first thing to determine is how many distinct sets of 4 (D+1) can be generated from the 23 (N) numbers. *For this example, that is COMBIN(N,D+1) = COMBIN(23,3+1) = 8,855 sets. *The second thing to determine is how many distinct sets of 4 (D+1) get used up every time you choose 7 (S) elements. *For this example, that is COMBIN(S,D+1) = COMBIN(7,3+1) = 35. *Therefore you can have 8,855 / 35 = 253 combinations. S Oops, sorry, I forgot to mention that this is a maximum. *The actual number of combinations the parameters permit when you attempt to construct the set may be lower. *I couldn't tell you off the top of my head how many instances or under what conditions it is less than the theoretical optimal number. An example of where it is less is using N=5, S=4, D=2, COMBIN(5,3)/ COMBIN(4,3) = 2.5. *When you attempt to construct the combinations, you'll find there is only 1. *(Any one of {1,2,3,4}, {1,2,3,5}, {1,2,4,5}, {1,3,4,5}, {1,2,3,4}, but you can't choose 2 of these as you'll end up violating the constraints) [In general, for N = k, S= k-1, and D=2, the actual number of sets is going to be 1.] You'll probably need someone with a bit more insight to help you determine when you will be unable to achieve the theoretical maximum. S- Hide quoted text - - Show quoted text - Thank you for your help. Gives me something to work with. |
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