nycjosh Posted November 4, 2010 Share Posted November 4, 2010 Okay.. I'm a complete idiot when it comes to math, but I have a question. I'm going try to my hardest to not make it sound like I barely graduated from a NYC public school... because that's what happened If I have four letters let say "abcd" how many different combinations of capitol letters can be placed in. For example "Abcd" "aBCd" "abcD" "AbcD" "aBcD" More so than knowing how many different combinations there are, is there a program that can give me the actual examples, kind of like a poker hand/odds calculator. Can I somehow get the actual examples/simulations, without writing the all out? Thanks in advance for any replies Josh Link to comment Share on other sites More sharing options...
ajb Posted November 4, 2010 Share Posted November 4, 2010 Just to be clear, you want to keep the canonical ordering "abcd"? For example, you would not include "Bacd"? (or rather it is equivalent to "aBcd"?) Link to comment Share on other sites More sharing options...
Shadow Posted November 4, 2010 Share Posted November 4, 2010 (edited) First, read up on the Rule of product. It is also possible to explain the answer using recursion, but I'm pretty sure the former will be easier to understand. So, we have four letters, and every letter can be in either of two states (capital, or not). Using the rule of product, we have [math] 2*2*2*2 = 2^4 = 16[/math] different combinations. If you want the number of combinations when you're allowed to swap the positions of letters (ie. BaCD), think of it this way: we have four places, and we want to place one of four letters in each spot in such a way that we only use every letter once. So, how many letters can we place in the first spot? Four of course. How many in the second? Three, because we already used one up in the first spot. Using this method and the rule of product, we have 4*3*2*1 = 4! = 24 ways of ordering the letters (4! is read as "four factorial", you can read about the factorial here). Now, again using the rule of product, we combine this result with the 16 ways to change the "capitalness" of the letters, which leaves us with the final answer, [math]24*16 = 384[/math]. Hope this helped. PS.: For some reason, putting "4!" in the math tags gives a LaTeX error, anyone know why? Edited November 4, 2010 by Shadow Link to comment Share on other sites More sharing options...
nycjosh Posted September 1, 2011 Author Share Posted September 1, 2011 Thanks guys! I've been meaning to reply but couldn't remember my password or email that I singed up with. I hate to bump such an old thread, but I think it's rude to not thank posters who give answers or help. Link to comment Share on other sites More sharing options...
khaled Posted September 2, 2011 Share Posted September 2, 2011 If we simplify the problem, by removing small letters assuming it's the other ones .. having a set of only the capitalized letters, by taking canonical order, we have set of size N, as combinations can be given: 0 ~ {} = {abcd} 1 ~ {A}, {B}, {C}, {D} = {Abcd}, {aBcd}, {abCd}, {abcD} 2 ~ {AB}, {BC}, {CD} = {ABcd}, {aBCd}, {abCD} 3 ~ {ABC}, {BCD} = {ABCd}, {aBCD} 4 ~ {ABCD} that's [math]1 + \sum{\underline N}[/math] Link to comment Share on other sites More sharing options...
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