MHB How to find set from following condition.

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The discussion revolves around finding sets based on specific conditions for the numbers {1, 2, 3}. The user seeks to understand how to calculate the number of possible combinations given that each number can exist in various states, with specific relational constraints among them. The initial confusion stems from the contradictory nature of the conditions, particularly regarding the fixed order of the numbers. Clarifications suggest that the user is interested in both the total number of sets and the methodology for determining these sets, especially for larger numbers. The conversation emphasizes the need for a clear problem statement to provide accurate solutions.
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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
 
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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.
 
rajemessage said:
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.
 
Ackbach said:
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
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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