how to find sets/combination for big numbers .

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rajemessage

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Aug 9, 2014
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Dear All,

how can i find the sets from following situation.
i have three numbers,{1 2 3} which will always be in this order {123},
i want to find out number of cases can be made,
.
but 2 can come at frist position that is before 1 or at second position or at
third position that is after 3.
and all are optional

for example power set of {123} is {1},{2},{23},{13},{12},{123}{3},{}.
now the problem is, for small numbers one can find out manually, but for big numbers , how can i find the arrangement/sets/combination of numbers.


any link will be helpfull.
yours sincerly
 
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Denis said:
Could you please repost in proper English, and provide an example. Thank you.
Or are you simply looking for:
123
132
213
231
312
321
Click to expand...


for example:- power set of {123} is {1},{2},{23},{13},{12},{123}{3},{}.
now the problem is, for small numbers one can find out manually, but for big numbers , how can i find the arrangement/sets/combination of numbers.
 
rajemessage said:
for example power set of {123} is {1},{2},{23},{13},{12},{123}{3},{}.
Click to expand...

The following uses bold for emphasis. Note that you must use commas to separate members of a set.


No, the power set of {123} is {{123}, {}}.


In contrast, the power set of {1, 2, 3} is {{1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}, {}}.
 
It's not really clear what you are asking. You say
i have three numbers,{1 2 3} which will always be in this order {123}
Click to expand...
but then you show 1, 2, and 3 in all possible orders. It looks to me like you are asking "How many subsets does a set {1, 2, 3, ..., n} have". That is well known- any set containing n elements has \(\displaystyle 2^n\) subsets. For example, in your example, {1, 2, 3}, there are 3 elements in the set so the number of subset, including the entire set and the empty set, is \(\displaystyle 2^3= 8\) and you have shown those 8 subsets in your post. If the set were {1, 2, 3, 4}, we could have each of the sets you show plus each of those sets with "4" appended. So the number of such subsets is 2 times the number of subsets of {1, 2, 3}: 2*8= 16. That's where the "2 to a power" comes from.
 
HallsofIvy said:
It's not really clear what you are asking. You say

but then you show 1, 2, and 3 in all possible orders. It looks to me like you are asking "How many subsets does a set {1, 2, 3, ..., n} have". That is well known- any set containing n elements has \(\displaystyle 2^n\) subsets. For example, in your example, {1, 2, 3}, there are 3 elements in the set so the number of subset, including the entire set and the empty set, is \(\displaystyle 2^3= 8\) and you have shown those 8 subsets in your post. If the set were {1, 2, 3, 4}, we could have each of the sets you show plus each of those sets with "4" appended. So the number of such subsets is 2 times the number of subsets of {1, 2, 3}: 2*8= 16. That's where the "2 to a power" comes from.
Click to expand...


yes, i have got my answer partially, that is, i have to append it at the end.

I have written only three elements at a time not four.
so i want the ans with this condition.
thirdly , i want to know formulas with which such questions can be formulated for big numbers.

like may be i can use binomial cofficeint or P & C formulas one by one for each situation. then find sets for them

"i do not know".

yours sincerley.
 
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