Log In
31 votes

A partial order $≤$ is defined on the set $S=\left \{ x, a_1, a_2, \ldots, a_n, y \right \}$ as $x$ $\leq _{i}$ $a_{i}$ for all $i$ and $a_{i}\leq y$ for all $i$, where $n ≥ 1$. The number of total orders on the set S which contain the partial order $≤$ is

  1. $n!$
  2. $n+2$
  3. $n$
  4. $1$
in Set Theory & Algebra
edited by

In question one typing mistake is that in place of "x <=i ai" , "x <= ai" should be written.

2 Answers

45 votes
Best answer

To make this partial order a total order, we need the relation to hold for every two element of the partial order. Currently between any $a_i$ and $a_j,$ there is no relation. So, for every $a_i, a_j,$ we have to add either $(a_i, a_j)$ or $(a_j, a_i)$ in total order. So, this translates to giving an ordering for $n$ elements between $x$ and $y,$ which can be done in $n!$ ways. So, answer is (a).

The bottom figure is for a total order. We can permute the $a_i$ from $i = 1$ to $n,$ and each permutation will also be a total order containing the given partial order. 

edited by
Answer to this is (C) i.e n. An order is set in which every two elements are comparable. So we can not take two or more than two ai in a TOSET as you have shown above. Tosets that are possible will be {x,a1,y},{x,a2,y}........{x,an,y}. So only 'n' tosets are possible.

To back up this argument we can also state a qoute from wikipedia that "maximum number of chains that an order can be divided into is ATMOST k, where k is the width of the lattice which here is n.
The question asks for NUMBER of total orders possible. {x, a1, a2, ... an, y} where there element at position i ≤ element at position i+1, is one such TOSET. Similarly, we can have {x, a2, a1, a3, ... an, y} which is another TOSET with the same relation. Both of these TOSET contains the given POSET in question. In this way we can permute in any order and each permutation will be a new TOSET and will contain the given POSET.
OK I got it..I was instead dividing the POSET into maximum number of TOSETS that it can be divided into with the same relation.

But question asks for TOSETs that contain the given POSET in it.

Thank You..
arjun sir,what is meant by the TOSET will contain the given poset?? how is it retained in the TOSET??Can you explain??
@Arjun. Its not that we have to order it in chain. Its that in how many ways we can have x1, x2......,xn arranged in between 'a' and 'b' so that the given partial order is preserved.
@Arjun Sir, I think the given ans fails for subset operation on Powerset of set A={a,b}.

where <= means subset operation


Here x=Phi a1={a} a2={b} y=A

we know {a} is not subset of {b} and vice versa. so both can't present in TOSET.

But u told in ur ans...a1 and a2 both can present simultaneously in n! order..which contradicts with above example.

let A={0,1,2,3,∞}

here x=0, y=∞, a1=1, a2=2, a3=3.

∞                                      ∞

|                                        |

1                and                 3

|                                        |

2                                       1

|                                         |

3                                       2

|                                        |

0                                       0

is a Toset                       is also a Toset 

bcz as mentiond in question , ≤ is applicable only for (x,a1) and (a3,y), however a1, a2, a(1,2,3) can be permutate in any order (3!=6) as long as x has leat value and y has greatest value in the set. same thing applicable for your query, i think. 


In question it is given that

x ≤ ai for all i and ai ≤ y for all i, 

  Rajesh Pradhan in ques they dont mention any relationship between a(i) and a(j) so u can arrange in any order

@Rajesh, Its mentioned in question about the partial order ≤ & analysis has to be done wrt that only. Dont change the Poset. Hope it clears your doubt.

Are we doing n! Permutations because relation between any ai and aj is not specified in question ???

We are doing n!, because in the question it is given that

The number of total orders on the set S which contain the partial order ≤ is

Cond 1: it should be TOSET:-

So, as per the definiition of the toset it should contain all the elements of the set,

 Cond 2: it should satisfy the relation x <= ai <= y.

That's why from cond 1 we have to take all n elements with x and y, and from cond 2 we can arrange in any order i.e. n!.

Hence, it is n! is the answer.

x is related to all a's and all a's are related to y in the partial order given. So x is the lowest element and y the top most. The a's can be arranged in any way which is n!

A partial order relation is total order , when it satisfies comparibility, means $a\leqslant b$ or $b\leqslant a$ . Then how poset looking like $a_{1},a_{2},.......$ has no relation between them? example

1 vote

The given POSET is drawn as shown in the best answer here. To make this into a TOSET, we have to make it a chain. So, connect all the a's together, and link one end of the "a chain" with x, and the other end with y.

The whole "a chain" can be made in $n!$ ways, and x and y are fixed.

So, possible Total orders = $n!$

Option A


Related questions

36 votes
4 answers
The number of equivalence relations of the set $\{1,2,3,4\}$ is $15$ $16$ $24$ $4$
asked Sep 29, 2014 in Set Theory & Algebra Kathleen 11.7k views
16 votes
6 answers
Let $\left(Z, *\right)$ be an algebraic structure where $Z$ is the set of integers and the operation $*$ is defined by $n*m = \max(n,m)$. Which of the following statements is true for $\left(Z, *\right)$? $\left(Z, *\right)$ is a monoid $\left(Z, *\right)$ is an Abelian group $\left(Z, *\right)$ is a group None of the above
asked Sep 29, 2014 in Set Theory & Algebra Kathleen 4.8k views
11 votes
3 answers
Let $R$ be a reflexive and transitive relation on a set $A$. Define a new relation $E$ on $A$ as $E=\{(a, b) \mid (a, b) \in R \text{ and } (b, a) \in R \}$ Prove that $E$ is an equivalence relation on $A$. Define a relation $\leq$ on the equivalence classes of $E$ as $E_1 \leq E_2$ if $\exists a, b$ such that $a \in E_1, b \in E_2 \text{ and } (a, b) \in R$. Prove that $\leq$ is a partial order.
asked Sep 29, 2014 in Set Theory & Algebra Kathleen 1.6k views
30 votes
3 answers
Let $F$ be the set of one-to-one functions from the set $\{1, 2, \dots, n\}$ to the set $\{1, 2,\dots, m\}$ where $m\geq n\geq1$. How many functions are members of $F$? How many functions $f$ in $F$ satisfy the property $f(i)=1$ for some $i, 1\leq i \leq n$? How many functions $f$ in $F$ satisfy the property $f(i)<f(j)$ for all $i,j \ \ 1\leq i \leq j \leq n$?
asked Sep 29, 2014 in Set Theory & Algebra Kathleen 3k views