how to find out combination from following situation

rajemessage

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Aug 9, 2014
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Dear All,


There are three numbers they can have two status "a" or "b".
except 1 , 1 will have only one state "a".
only one status can exist at one time for a number.
I have to chose three number(1,2,3) , they will be in this order only.
, so please tel how many combination
can be there?


1(a)
2(a)
2(b)
3(a)
3(b)


yours sincerley

new formulation
--------------------
It is bit different,with your permition,i am taking this apartunity to make some changes.
1 2 3 will always be in this order {123}, i want to find out number of cases can be made,
like {1},{2},{23},{13},{12},{123}{3},{}. but each number has two states like "a" "b", i.e, each one will become different entity,like 2a,2b,3a,3b,1a,
with only exception i.e. 1 will have only one state 1a.

please tel me step wise using formulas, so that i can understand, also any link will be helpfull.
yours sincerly
 
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It's a little hard to understand exactly what you are asking- especially that list at the end!

I think you are saying that you have three number of which the first is 1 and the next two can be either 1 or 2. Since the second number can have two values, and the third number can have two values, the combination can have 2*2= 4 values- they are 11, 12, 21, and 22. The initial "1", since it is fixed, is irrelevant to the number of such combinations- there are 4 such combinations of three numbers, 111, 112, 121, and 122.
 
It's a little hard to understand exactly what you are asking- especially that list at the end!

I think you are saying that you have three number of which the first is 1 and the next two can be either 1 or 2. Since the second number can have two values, and the third number can have two values, the combination can have 2*2= 4 values- they are 11, 12, 21, and 22. The initial "1", since it is fixed, is irrelevant to the number of such combinations- there are 4 such combinations of three numbers, 111, 112, 121, and 122.


It is bit different,with your permition,i am taking this apartunity to make some changes.
1 2 3 will always be in this order {123}, i want to find out number of cases can be made,
like {1},{2},{23},{13},{12},{123}{3},{}. but each number has two states like "a" "b", i.e, each one will become different entity,like 2a,2b,3a,3b,1a,
with only exception i.e. 1 will have only one state 1a.

please tel me step wise using formulas, so that i can understand, also any link will be helpfull.
yours sincerly
 
I am not sure what you mean by a number having "two states". It sounds like you mean you have six things, 1a, 1b, 2a, 2b, 3a, 3b but that "1a" and "1b" cannot appear together. There would be \(\displaystyle 2^3= 8\) subsets of a set of 3 things, 7 of which are non-empty. Adding that each item can have "two states" doubles all of the non-empty sets giving 14+ 1= 15such sets.
 
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Sounds to me like the problem is, how many numbers are there which start with the number 1, have a maximum number of digits of 3, and each digit after the first one (if there is one) is either 1, 2 or 3.

Given that, we also need to know whether, for the next choice of a digit, you can replace the digit you choose after choosing, i.e. (1) each digit can be used multiple times so that 111 is allowed, or (2) you don't replace the digits, i.e. each digit can only be used once and 111 is not allowed.

In each case you have 3 sub cases: (a) a length of 1, (b) 2, and (c) 3. Since case (a) is always just the single number 1, just figure out the cases for (b), and (c) and add 1. The general case where you are allowed to choose a length of zero (no digit is chosen), is not allowed because the number must start with 1, i.e. you must choose 1 for the first digit.

Case 1:
(b) 3 choices = 3 numbers [11, 12 and 13]
(c) 3 choices times 3 choices = 9 numbers [111, 112, 113, 121, 122, 123, 131, 132, 133]
So (3 + 9) + 1 = choices

In general this case gives sum of xn for n = 1, 2, 3, ..., N-1 minus 1 where x is the number of digits (in this case 3) and N is the number of unique digits allowed (in this case 3). The formula for this is [xN -1]/[x-1] - 1

Hopefully the example will help you figure out case (2) which I think you meant, but let us know.
 
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I am not sure what you mean by a number having "two states". It sounds like you mean you have six things, 1a, 1b, 2a, 2b, 3a, 3b but that "1a" and "1b" cannot appear together. There would be \(\displaystyle 2^3= 8\) subsets of a set of 3 things, 7 of which are non-empty. Adding that each item can have "two states" doubles all of the non-empty sets giving 14+ 1= 15such sets.

a condition is not full filled, 1 will have only "1a".
"1b" does not exist.

yours sincerly
 
I still do not understand what you mean by "have 1a". However, I was wrong- I forgot about "2b" and "2c".
The original 8 sets involving only 1a, 1b, and 1c are
{} {1a} {1b} {1c} {1a, 1b} {1a, 1c} {1b, 1c}, {1a, 1b, 1c}
then, swapping 2a for 1a
{2a} {2a, 1b} {2a, 1c} {1b, 1c}, {2a, 1b, 1c}
only 5 new sets since 3 of the original 8 did not contain 1a.
then, swapping 2b for 2a
{2b}{1a, 2b} {1a, 1c} {2b, 1c}, {1a, 2b, 1c}
again 5 new sets.
then, swapping 2c for 2a:
{2c} {1a, 2b} {1a, 2c} {1b, 2c}, {1a, 1b, 2c}
again 5 new sets.
There were only two sets, {1a, 1b1} and {1a, 1b, 1c} that contain both 1a and 1b in the first 8 so we can get two new sets by replacing 1a and 1b with 2a and 2b
{2a, 2b} {2a, 2b, 1c}
There were two sets, {1a, 1c} and {1a, 1b, 1c} in the first 8 that contain both 1a and 1c so we can get two new sets
{2a, 2c} {2a, 1b, 2c}
There were two sets, {1b, 1c} and {1a, 1b, 1c} in the first 8 that contain both 1b and 1c so we can get two new sets
{2b, 2c} {1a, 2b, 2c}
Finally, there was one set {1a, 1b, 1c} that contained all three so we have the new set
{2a, 2b, 2c}

That is a total of 8+ 3(5)+ 3(2)+ 1= 8+ 15+ 6+ 1= 30 such sets.

That is my interpretation of what you were trying to do- finding the number of sets if you have 1a, 2a, 1b, 2b, 1c, 2c but the two "a"s cannot be in the same set, the two "b"s cannot be in the same set, and the two "c"s cannot be in the same set. If that is not correct please explain exactly what you mean.
 
Dear All,


There are three numbers they can have two status "a" or "b".
except 1 , 1 will have only one state "a".
only one status can exist at one time for a number.
I have to chose three number(1,2,3) , they will be in this order only.
, so please tel how many combination
can be there?


1(a)
2(a)
2(b)
3(a)
3(b)


yours sincerley

Another interpretation: There are three numbers (1, 2, and 3)
(1) The number 1 can only have one state and that is state 1a.
(2) The number 2 can have two states and those are states 2a and 2b.
(3) The number 3 can have two states and those are states 3a and 3b.

How many lists (subsets) of the set of different combinations of these five states can be made such that, if they are in the subset, the state 1a is listed first, the states of 2 are listed second in order 2a and 2b, and the states of 3 are listed third in order 3a and 3b.

Once again this breaks down into a number of sub-cases, those of length L for L=0, 1, 2, 3, 4, 5. We can count the number of cases in each set, i.e.
Length 0; 1 subset - the empty set {}.
Length 1; 5 subsets - {1a}, {2a}, {2b}, {3a}, {3b}
...
and develop a general formula but it isn't worth it unless this is the proper interpretation of the problem.
 
Another interpretation: There are three numbers (1, 2, and 3)
(1) The number 1 can only have one state and that is state 1a.
(2) The number 2 can have two states and those are states 2a and 2b.
(3) The number 3 can have two states and those are states 3a and 3b.

How many lists (subsets) of the set of different combinations of these five states can be made such that, if they are in the subset, the state 1a is listed first, the states of 2 are listed second in order 2a and 2b, and the states of 3 are listed third in order 3a and 3b.

Once again this breaks down into a number of sub-cases, those of length L for L=0, 1, 2, 3, 4, 5. We can count the number of cases in each set, i.e.
Length 0; 1 subset - the empty set {}.
Length 1; 5 subsets - {1a}, {2a}, {2b}, {3a}, {3b}
...
and develop a general formula but it isn't worth it unless this is the proper interpretation of the problem.

thanku, for big numbers it is quite tedious to find the sets, is there any way to find them.
yours sincerley
 
thanku, for big numbers it is quite tedious to find the sets, is there any way to find them.
yours sincerley

Yes, you're right, for big numbers it is quite tedious. You might develop a fuller understanding of combinations and permutations if you were to read
http://www.mathsisfun.com/combinatorics/combinations-permutations.html
That page develops formulas for each of the length 0, length 1, length 2, ... groups depending on what assumption you make about order being important or not and whether replacement is allowed or not [for example, you might want to pose the probem to allow any of the states to occur more than once].
 
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