How to find set from following condition.

[rajemessage]

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.

please solve this question with forumulas so that i can find set of bignumbers too.

yours sincerly

 

[MarkFL]

If you have $n$ objects, then there are $n!$ ways to order them. You have $n$ choices for the first position, $n-1$ for the second and so on.

 

[Ackbach]

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.

please solve this question with forumulas so that i can find set of bignumbers too.

yours sincerly

Your problem statement is self-contradictory. The numbers cannot always be in the order $\{1,2,3\}$ if you're allowing $2$ to be at the first position. Or are you allowing the order to "wrap around"? In that case, you could have $\{1,2,3\}, \{2,3,1\}, \{3,1,2\}$, so there are $n$ possibilities.

Or are you asking how many permutations of the set there are? If so, MarkFL's answer is correct.

Or are you asking how many subsets of $\{1,2,3\}$ there are (or what is the power set)? If so, then think about for any given subset, whether you're going to include an element or not, and that'll get you going in the right direction.

It would be very helpful if you could please give us the original problem statement, word-for-word.

 

[rajemessage]

Your problem statement is self-contradictory. The numbers cannot always be in the order $\{1,2,3\}$ if you're allowing $2$ to be at the first position. Or are you allowing the order to "wrap around"? In that case, you could have $\{1,2,3\}, \{2,3,1\}, \{3,1,2\}$, so there are $n$ possibilities.

Or are you asking how many permutations of the set there are? If so, MarkFL's answer is correct.

Or are you asking how many subsets of $\{1,2,3\}$ there are (or what is the power set)? If so, then think about for any given subset, whether you're going to include an element or not, and that'll get you going in the right direction.

It would be very helpful if you could please give us the original problem statement, word-for-word.

yes 123 can not be in this order if i change the position of 2, basically it is my condition.

i am re writing the same problem with conditions.

i want to find out set/case in following situation.

there are three number.{1,2,3}

"1" can have three states,(it will not exist,it will exist and it will be always less than "3")

"2" can have four states, it can be less than "1" , greater than "1" and less than "3", greater than "3" ,existing ,not existing.

"3" can have three states,(not exist,exist and it will be always greater than "1")

"all numbers 1,2,3 are optional".

Q1) how many sets will be there
Q2) what will be that sets/combination
Q3) how to find condition based sets for big number like above

yours sincerely

 

[CellCoree]

,I would approach this problem by first defining what a set is and how it relates to the given situation. A set is a collection of distinct elements, which in this case would be the numbers 1, 2, and 3. The order of these elements does not matter in a set, so {1,2,3} is the same as {2,1,3}.

Based on the given information, there are three possible positions for the number 2: before 1, after 3, or between 1 and 3. This means that there are three possible combinations for the set: {1,2,3}, {2,1,3}, and {1,3,2}. It is important to note that {2,3,1} and {3,2,1} are not considered separate sets because they have the same elements in a different order.

To find the total number of possible cases, we can use the formula for combinations, which is nCr = n! / (r! * (n-r)!). In this case, n is equal to 3 because there are 3 numbers and r is equal to 3 because we want to find all possible combinations. Plugging these values into the formula, we get 3! / (3! * (3-3)!) = 3! / (3! * 0!) = 3! / 3! = 1. This means that there is only one possible set in this situation, which is {1,2,3}.

If we want to extend this problem to find sets with bigger numbers, we can use the same approach. For example, if we have the numbers {1,2,3,4}, there would be 4! / (4! * (4-4)!) = 4! / 4! = 1 possible set, which is {1,2,3,4}.

In conclusion, the number of possible sets that can be made from the given situation is 1, and the formula for finding this value is nCr = n! / (r! * (n-r)!). This approach can be used to find sets with any number of elements.