You are not logged in.
Pages: 1
Hello. With the wide variety of information around the internet there are still some specific stuffs that can't be answered. Thus I am here to ask about it you the experts. Here is the problem.
There are 4 letters A B C and D. Each letter have 3 variety of color Red, Yellow and Green. So basically the letters goes as red A, yellow A, green A, red B, yellow B, green B and so on and so forth. How many permutations can there be if each permutation should have only 3 letters which is different from each other. Example; (Red A, Red B, Red C), (Red A, Red B, Red D), (Yellow A, Red C, Green D) and so on..
I tried solving it manually but its getting a bit more confusing than I expected. What I would humbly ask is the complete arrangement of the letters.
Thank you very much.
Last edited by Jhaesen (2016-05-27 12:13:44)
Offline
Hi;
Question 1:
Do you require the number or the actual list?
Question 2:
You said permutations but I will still check on this point. Do you consider
(Yellow A, Red C, Green D) and (Red C, Green D, Yellow A) as different or the same?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
Answer 1. If I can get the actual list, I'll be very grateful. But I am currently in a middle of making an excel generated permutations for this issue. I just need to find the exact formula to make it work properly.
Answer 2. Yes, I refer it as permutations because (yellow A, Red C, Green D) is different from (Red C, Green D, Yellow A).
Offline
{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
Offline
Sir, can I humbly ask you to explain your answer so I can code it properly in excel. Thank you very much for your time.
Last edited by Jhaesen (2016-05-27 15:29:48)
Offline
Hello: I think that answer is too much. If you simulate it you will agree.
Offline
Hi;
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
Hi;
Thank you very much Sir. I also did some research while waiting for some response and this is also what I came up with me research.
Offline
First we consider the permutations of simply the letters A,B,C and D. Arranging 3 letters out of 4 is
As far as colors are concerned there are 4 colors. So each letter can have 4 mutually exclusive patterns . so it is multiplied by 4*4*4 i.e. So totally{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
Offline
Hi;
You can pick any one of 12 cards first then you can pick any one of 9 remaining second and then you have 6 left to pick for the third.
12 * 9 * 6 = 648
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
Last edited by thickhead (2016-05-27 18:09:54)
{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
Offline
Yep!
EVW;
Here is the M code you asked for.
types = Flatten[
Table[{color,
letter}, {color, {"red", "yellow", "green"}}, {letter , {"A",
"B", "C", "D"}}], 1];
s = Permutations[types, {3}];
ans = Cases[s, {{a_, b_}, {c_, d_}, {e_, f_}} /; b != d != f] //
Length
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
Pages: 1