jumpergirl 0 #1 January 17, 2006 What is the formula for figuring how many possible combinations are available? I have a model home with 5 options. You can chose and combination you want (from no options to all 5). How do I determine how many possible combinations I can have? Thanks! Quote Share this post Link to post Share on other sites waltappel 1 #2 January 17, 2006 QuoteWhat is the formula for figuring how many possible combinations are available? I have a model home with 5 options. You can chose and combination you want (from no options to all 5). How do I determine how many possible combinations I can have? Thanks! http://mathforum.org/dr.math/faq/faq.comb.perm.html Walt Quote Share this post Link to post Share on other sites NtheSeaOrSky 0 #3 January 17, 2006 5^5 (five to the fifth power)Life is not fair and there are no guarantees... Quote Share this post Link to post Share on other sites jumpergirl 0 #4 January 17, 2006 I found that site, too, but I don't get it. I remember learning this in high school, but I can't "read the language" anymore. Can you translate that to normal English, please? Quote Share this post Link to post Share on other sites jumpergirl 0 #5 January 17, 2006 I feel so stupid right now. What does 5^5 mean? Is it 5*5*5*5*5? Or 5*5? I need to go back to high school, I guess. Quote Share this post Link to post Share on other sites NtheSeaOrSky 0 #6 January 17, 2006 sorry 6 to the fifth power 5 options ___ ___ ____ ___ ___ and you can choose 6 for each (0,1,2,3,4,5) so 6x6x6x6x6Life is not fair and there are no guarantees... Quote Share this post Link to post Share on other sites wildblue 7 #7 January 17, 2006 Unless I haven't had enough coffee this morning, it sounds like you'll have to add up the possible combinations (5C0, 5C1, 5C2, ...)it's like incest - you're substituting convenience for quality Quote Share this post Link to post Share on other sites jumpergirl 0 #8 January 17, 2006 That's too high. I forgot to mention, there can't be any duplicates. You have Model B and can choose any or all options, but can't have 2 of the same. Model B- Wap Patio Porch Veranda 5th bedroom Does that make a difference? Quote Share this post Link to post Share on other sites NtheSeaOrSky 0 #9 January 17, 2006 LOL, yes it makes a difference lol ok so 5 options ___ ___ ___ ___ ___ 6x5x4x3x2 (each option you are taking away a possibility with no duplicates)Life is not fair and there are no guarantees... Quote Share this post Link to post Share on other sites jumpergirl 0 #10 January 17, 2006 That's 720. I think that's too many. I have come up with 27 on my own, by doing as WildBlue said, and writing each one seperate. UGH! Quote Share this post Link to post Share on other sites NtheSeaOrSky 0 #11 January 17, 2006 well go with yours then, I might be misunderstanding what you are asking.Life is not fair and there are no guarantees... Quote Share this post Link to post Share on other sites waltappel 1 #12 January 17, 2006 QuoteI found that site, too, but I don't get it. I remember learning this in high school, but I can't "read the language" anymore. Can you translate that to normal English, please? I'm about to go into a meeting. As soon as it's over, I'll do that. Walt Quote Share this post Link to post Share on other sites wildblue 7 #13 January 17, 2006 n! n_C_k = ---------- k!(n - k)! n = number of things k = number choosen So I'd add 5C0 + 5C1 + 5C2 + ... edit in case you forgot, ! means 'factorial' so 5! = 5*4*3*2*1 WAG, 37 or 42... I think.it's like incest - you're substituting convenience for quality Quote Share this post Link to post Share on other sites rehmwa 2 #14 January 17, 2006 QuoteWhat is the formula for figuring how many possible combinations are available? I have a model home with 5 options. You can chose and combination you want (from no options to all 5). How do I determine how many possible combinations I can have? Thanks! 2 raised to the 5th = 32 combinations, including a null option of none of them (as long as the options are completely independent and only yes/no type of option) Each option is yes/no - only 2 choices so 2 options is 2x2 3 options is (2x2)x2 (add the 3rd option) and so on ... Driving is a one dimensional activity - a monkey can do it - being proud of your driving abilities is like being proud of being able to put on pants Quote Share this post Link to post Share on other sites wildblue 7 #15 January 17, 2006 Quote2 raised to the 5th = 32 combinations Leave it to me to over complicate a simple problem! it's like incest - you're substituting convenience for quality Quote Share this post Link to post Share on other sites tso-d_chris 0 #16 January 17, 2006 QuoteQuoteWhat is the formula for figuring how many possible combinations are available? I have a model home with 5 options. You can chose and combination you want (from no options to all 5). How do I determine how many possible combinations I can have? Thanks! 2 raised to the 5th = 32 combinations, including a null option of none of them (as long as the options are completely independent and only yes/no type of option) Each option is yes/no - only 2 choices so 2 options is 2x2 3 options is (2x2)x2 (add the 3rd option) and so on We have a winner! For Great Deals on Gear Quote Share this post Link to post Share on other sites Kimblair13 0 #17 January 17, 2006 nCr = n!/r!(n-r)! If you have a scientific calculator it as a nCr button. Quote Share this post Link to post Share on other sites waltappel 1 #18 January 17, 2006 QuoteWhat is the formula for figuring how many possible combinations are available? I have a model home with 5 options. You can chose and combination you want (from no options to all 5). How do I determine how many possible combinations I can have? Thanks! Ok. Here's the deal. When you are talking taking some number of objects from a group without regard to order, that is called a combination. What you have a group of five options, but you can choose any number of them or none at all. Lets's say you choose none of the options, there is only 1 way to do that. If you take only one option, there are 5 choices. If you take two options, there are 10 ways to do that. and so on. It all adds up to 1 + 5 + 10 + 10 + 5 + 1 = 32 Walt Quote Share this post Link to post Share on other sites AshDeBash 0 #19 January 17, 2006 If you can count in binary, it might make it easier to understand why your solution is 2^5 (2*2*2*2*2) Think of 0=option not required and 1=Option required. Then assign each of the options into a column. Therefore... 00000 (no options) 00001 (only option assigned to column 5 is required) 00010 00011 00100 00101 00110 00111 (options 3, 4 and 5 required etc., etc.) 01000 01001 01010 01011 01100 01101 01110 01111 10000 10001 10010 10011 10100 10101 10110 10111 11000 11001 11010 11011 11100 11101 11110 11111 Hence 32 combinations. Blimey - I think I'm a geek! Quote Share this post Link to post Share on other sites tso-d_chris 0 #20 January 17, 2006 QuoteQuoteWhat is the formula for figuring how many possible combinations are available? I have a model home with 5 options. You can chose and combination you want (from no options to all 5). How do I determine how many possible combinations I can have? Thanks! Ok. Here's the deal. When you are talking taking some number of objects from a group without regard to order, that is called a combination. What you have a group of five options, but you can choose any number of them or none at all. Lets's say you choose none of the options, there is only 1 way to do that. If you take only one option, there are 5 choices. If you take two options, there are 10 ways to do that. and so on. It all adds up to 1 + 5 + 10 + 10 + 5 + 1 = 32 Walt 5 options --> 2^5 = 32 For Great Deals on Gear Quote Share this post Link to post Share on other sites Superman32 0 #21 January 17, 2006 I've always liked the idea of factorials but I too thought it was used to find a number of possibilities, ie, if I had five individual number and I would take the factorial of 5 to find the number of possible combinations. is this right? Somebody please enlighten me on one. Inveniam Viam aut Faciam I'm back biatches! Quote Share this post Link to post Share on other sites Superman32 0 #22 January 17, 2006 Abnother question: If I had a lock with for distinct rows and 10 numbers each, how many combination possibilities are there? Inveniam Viam aut Faciam I'm back biatches! Quote Share this post Link to post Share on other sites rehmwa 2 #23 January 17, 2006 QuoteAbnother question: If I had a lock with for distinct rows and 10 numbers each, how many combination possibilities are there? 10 choices for each combination - 10x10x10x10 (the lock counts from 0000 up to 9999) if it was letters, it would be 26x26x26x26 if it was 2 letters and 2 numbers, then 26x10x26x10 and so on - that should answer any similar question going forward ... Driving is a one dimensional activity - a monkey can do it - being proud of your driving abilities is like being proud of being able to put on pants Quote Share this post Link to post Share on other sites tso-d_chris 0 #24 January 17, 2006 QuoteI've always liked the idea of factorials but I too thought it was used to find a number of possibilities, ie, if I had five individual number and I would take the factorial of 5 to find the number of possible combinations. is this right? Somebody please enlighten me on one. In this case, each option has two possibilities, and choosing one option does not preclude other options being chosen, if I understood the original problem correctly. So each option has two possibilities, with a total of five options, which works out to 2^5=32. If each option had three possibilities, the total number of possibilities would be 3^5. With four possibilities, the answer would be 4^5. Now for the quiz : How high can one count on their hands, assuming both hands are used, and they each have five fingers? For Great Deals on Gear Quote Share this post Link to post Share on other sites mark 107 #25 January 17, 2006 Quote5 options --> 2^5 = 32 32 is correct if you assume each of the options is independent, and each option can have only two values. Some options may have more than two values. Paint color, for example. Or paint finish: matte, eggshell, semi-gloss, gloss, satin. Some options may be dependent on others. For example, one style of cabinet door may be available only in white laminate, while another may be available in oak, maple, or cherry. There's not enough given information for a complete answer. Mark Quote Share this post Link to post Share on other sites Prev 1 2 Next Page 1 of 2 Join the conversation You can post now and register later. If you have an account, sign in now to post with your account. Note: Your post will require moderator approval before it will be visible. Reply to this topic... × Pasted as rich text. Paste as plain text instead Only 75 emoji are allowed. × Your link has been automatically embedded. Display as a link instead × Your previous content has been restored. Clear editor × You cannot paste images directly. Upload or insert images from URL. Insert image from URL × Desktop Tablet Phone Submit Reply 0
waltappel 1 #2 January 17, 2006 QuoteWhat is the formula for figuring how many possible combinations are available? I have a model home with 5 options. You can chose and combination you want (from no options to all 5). How do I determine how many possible combinations I can have? Thanks! http://mathforum.org/dr.math/faq/faq.comb.perm.html Walt Quote Share this post Link to post Share on other sites
NtheSeaOrSky 0 #3 January 17, 2006 5^5 (five to the fifth power)Life is not fair and there are no guarantees... Quote Share this post Link to post Share on other sites
jumpergirl 0 #4 January 17, 2006 I found that site, too, but I don't get it. I remember learning this in high school, but I can't "read the language" anymore. Can you translate that to normal English, please? Quote Share this post Link to post Share on other sites
jumpergirl 0 #5 January 17, 2006 I feel so stupid right now. What does 5^5 mean? Is it 5*5*5*5*5? Or 5*5? I need to go back to high school, I guess. Quote Share this post Link to post Share on other sites
NtheSeaOrSky 0 #6 January 17, 2006 sorry 6 to the fifth power 5 options ___ ___ ____ ___ ___ and you can choose 6 for each (0,1,2,3,4,5) so 6x6x6x6x6Life is not fair and there are no guarantees... Quote Share this post Link to post Share on other sites
wildblue 7 #7 January 17, 2006 Unless I haven't had enough coffee this morning, it sounds like you'll have to add up the possible combinations (5C0, 5C1, 5C2, ...)it's like incest - you're substituting convenience for quality Quote Share this post Link to post Share on other sites
jumpergirl 0 #8 January 17, 2006 That's too high. I forgot to mention, there can't be any duplicates. You have Model B and can choose any or all options, but can't have 2 of the same. Model B- Wap Patio Porch Veranda 5th bedroom Does that make a difference? Quote Share this post Link to post Share on other sites
NtheSeaOrSky 0 #9 January 17, 2006 LOL, yes it makes a difference lol ok so 5 options ___ ___ ___ ___ ___ 6x5x4x3x2 (each option you are taking away a possibility with no duplicates)Life is not fair and there are no guarantees... Quote Share this post Link to post Share on other sites
jumpergirl 0 #10 January 17, 2006 That's 720. I think that's too many. I have come up with 27 on my own, by doing as WildBlue said, and writing each one seperate. UGH! Quote Share this post Link to post Share on other sites
NtheSeaOrSky 0 #11 January 17, 2006 well go with yours then, I might be misunderstanding what you are asking.Life is not fair and there are no guarantees... Quote Share this post Link to post Share on other sites
waltappel 1 #12 January 17, 2006 QuoteI found that site, too, but I don't get it. I remember learning this in high school, but I can't "read the language" anymore. Can you translate that to normal English, please? I'm about to go into a meeting. As soon as it's over, I'll do that. Walt Quote Share this post Link to post Share on other sites
wildblue 7 #13 January 17, 2006 n! n_C_k = ---------- k!(n - k)! n = number of things k = number choosen So I'd add 5C0 + 5C1 + 5C2 + ... edit in case you forgot, ! means 'factorial' so 5! = 5*4*3*2*1 WAG, 37 or 42... I think.it's like incest - you're substituting convenience for quality Quote Share this post Link to post Share on other sites
rehmwa 2 #14 January 17, 2006 QuoteWhat is the formula for figuring how many possible combinations are available? I have a model home with 5 options. You can chose and combination you want (from no options to all 5). How do I determine how many possible combinations I can have? Thanks! 2 raised to the 5th = 32 combinations, including a null option of none of them (as long as the options are completely independent and only yes/no type of option) Each option is yes/no - only 2 choices so 2 options is 2x2 3 options is (2x2)x2 (add the 3rd option) and so on ... Driving is a one dimensional activity - a monkey can do it - being proud of your driving abilities is like being proud of being able to put on pants Quote Share this post Link to post Share on other sites
wildblue 7 #15 January 17, 2006 Quote2 raised to the 5th = 32 combinations Leave it to me to over complicate a simple problem! it's like incest - you're substituting convenience for quality Quote Share this post Link to post Share on other sites
tso-d_chris 0 #16 January 17, 2006 QuoteQuoteWhat is the formula for figuring how many possible combinations are available? I have a model home with 5 options. You can chose and combination you want (from no options to all 5). How do I determine how many possible combinations I can have? Thanks! 2 raised to the 5th = 32 combinations, including a null option of none of them (as long as the options are completely independent and only yes/no type of option) Each option is yes/no - only 2 choices so 2 options is 2x2 3 options is (2x2)x2 (add the 3rd option) and so on We have a winner! For Great Deals on Gear Quote Share this post Link to post Share on other sites
Kimblair13 0 #17 January 17, 2006 nCr = n!/r!(n-r)! If you have a scientific calculator it as a nCr button. Quote Share this post Link to post Share on other sites
waltappel 1 #18 January 17, 2006 QuoteWhat is the formula for figuring how many possible combinations are available? I have a model home with 5 options. You can chose and combination you want (from no options to all 5). How do I determine how many possible combinations I can have? Thanks! Ok. Here's the deal. When you are talking taking some number of objects from a group without regard to order, that is called a combination. What you have a group of five options, but you can choose any number of them or none at all. Lets's say you choose none of the options, there is only 1 way to do that. If you take only one option, there are 5 choices. If you take two options, there are 10 ways to do that. and so on. It all adds up to 1 + 5 + 10 + 10 + 5 + 1 = 32 Walt Quote Share this post Link to post Share on other sites
AshDeBash 0 #19 January 17, 2006 If you can count in binary, it might make it easier to understand why your solution is 2^5 (2*2*2*2*2) Think of 0=option not required and 1=Option required. Then assign each of the options into a column. Therefore... 00000 (no options) 00001 (only option assigned to column 5 is required) 00010 00011 00100 00101 00110 00111 (options 3, 4 and 5 required etc., etc.) 01000 01001 01010 01011 01100 01101 01110 01111 10000 10001 10010 10011 10100 10101 10110 10111 11000 11001 11010 11011 11100 11101 11110 11111 Hence 32 combinations. Blimey - I think I'm a geek! Quote Share this post Link to post Share on other sites
tso-d_chris 0 #20 January 17, 2006 QuoteQuoteWhat is the formula for figuring how many possible combinations are available? I have a model home with 5 options. You can chose and combination you want (from no options to all 5). How do I determine how many possible combinations I can have? Thanks! Ok. Here's the deal. When you are talking taking some number of objects from a group without regard to order, that is called a combination. What you have a group of five options, but you can choose any number of them or none at all. Lets's say you choose none of the options, there is only 1 way to do that. If you take only one option, there are 5 choices. If you take two options, there are 10 ways to do that. and so on. It all adds up to 1 + 5 + 10 + 10 + 5 + 1 = 32 Walt 5 options --> 2^5 = 32 For Great Deals on Gear Quote Share this post Link to post Share on other sites
Superman32 0 #21 January 17, 2006 I've always liked the idea of factorials but I too thought it was used to find a number of possibilities, ie, if I had five individual number and I would take the factorial of 5 to find the number of possible combinations. is this right? Somebody please enlighten me on one. Inveniam Viam aut Faciam I'm back biatches! Quote Share this post Link to post Share on other sites
Superman32 0 #22 January 17, 2006 Abnother question: If I had a lock with for distinct rows and 10 numbers each, how many combination possibilities are there? Inveniam Viam aut Faciam I'm back biatches! Quote Share this post Link to post Share on other sites
rehmwa 2 #23 January 17, 2006 QuoteAbnother question: If I had a lock with for distinct rows and 10 numbers each, how many combination possibilities are there? 10 choices for each combination - 10x10x10x10 (the lock counts from 0000 up to 9999) if it was letters, it would be 26x26x26x26 if it was 2 letters and 2 numbers, then 26x10x26x10 and so on - that should answer any similar question going forward ... Driving is a one dimensional activity - a monkey can do it - being proud of your driving abilities is like being proud of being able to put on pants Quote Share this post Link to post Share on other sites
tso-d_chris 0 #24 January 17, 2006 QuoteI've always liked the idea of factorials but I too thought it was used to find a number of possibilities, ie, if I had five individual number and I would take the factorial of 5 to find the number of possible combinations. is this right? Somebody please enlighten me on one. In this case, each option has two possibilities, and choosing one option does not preclude other options being chosen, if I understood the original problem correctly. So each option has two possibilities, with a total of five options, which works out to 2^5=32. If each option had three possibilities, the total number of possibilities would be 3^5. With four possibilities, the answer would be 4^5. Now for the quiz : How high can one count on their hands, assuming both hands are used, and they each have five fingers? For Great Deals on Gear Quote Share this post Link to post Share on other sites
mark 107 #25 January 17, 2006 Quote5 options --> 2^5 = 32 32 is correct if you assume each of the options is independent, and each option can have only two values. Some options may have more than two values. Paint color, for example. Or paint finish: matte, eggshell, semi-gloss, gloss, satin. Some options may be dependent on others. For example, one style of cabinet door may be available only in white laminate, while another may be available in oak, maple, or cherry. There's not enough given information for a complete answer. Mark Quote Share this post Link to post Share on other sites
