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A partial order ≤ is defined on the set S = {x, a1, a2, ... an, y} as x ≤ ai for all i and ai ≤ 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

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 ai and aj, there is no relation. So, for every ai, aj, we have to add either (ai, aj) or (aj, ai) 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 ai from i = 1 to n, and each permutation will also be a total order containing the given partial order.

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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 ???