Relations and Combinatorics Question

[Krishang]

Let there be a set A such that [imath] A = {1,2,3,...,2026} [/imath]
Then find the number of relations on set A such that it's reflexive, symmetric and has exactly (1013)² + (1013) - 2 elements.

The answer is [imath] \binom{2(1013)²-1013}{\frac{(1013)²-1013-2}{2}} [/imath].

Maybe we have to first include all the 2026 elements of the form [imath] (a,a) [/imath] and also the other ones which are reflexive so we will have [imath] \frac{\binom{2026}{2} - 2026}{2} [/imath] to choose from but then how should I proceed, if this is the right approach?

Thank you.

 

[nasi112]

You will run out of fuel if you start with a set that contain 2026 elements. You should start low and look for a pattern. This is what I do.
Also for clarity you should start with definitions.
Reflexive Relation
A relation [imath]R[/imath] on a set [imath]A[/imath] is reflexive if [imath](a,a) \in R [/imath] for every [imath]a \in A[/imath].
Symmetric Relation
A relation [imath]R[/imath] on a set [imath]A[/imath] is symmetric if [imath](a,b) \in R [/imath], then [imath](b,a) \in R [/imath] for all [imath]a,b \in A[/imath].
Read the definitions carefully. If you understand them, you can start with a set with one element only.
[imath]A = \{1\}[/imath]
[imath]R = \{(1,1)\}[/imath]
One relation.
If [imath]A = \{1,2\}[/imath]
[imath]R_1 = \{(1,1),(2,2)\}[/imath]
[imath]R_2 = \{(1,1),(2,2),(1,2),(2,1)\}[/imath]
Two relations.
If [imath]A = \{1,2,3\}[/imath]
[imath]R_1 = \{(1,1),(2,2),(3,3)\}[/imath]
[imath]R_2 = \{(1,1),(2,2),(3,3),(1,2),(2,1)\}[/imath]
[imath]R_3 = \{(1,1),(2,2),(3,3),(1,3),(3,1)\}[/imath]
[imath]R_4 = \{(1,1),(2,2),(3,3),(2,3),(3,2)\}[/imath]
[imath]R_5 = \{(1,1),(2,2),(3,3),(1,2),(2,1),(1,3),(3,1)\}[/imath]
[imath]R_6 = \{(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)\}[/imath]
[imath]R_7 = \{(1,1),(2,2),(3,3),(2,3),(3,2),(1,3),(3,1)\}[/imath]
[imath]R_8 = \{(1,1),(2,2),(3,3),(1,2),(2,1),(1,3),(3,1),(2,3),(3,2)\}[/imath]
Eight relations.
Do you see the pattern?
[imath]2^0 = 1[/imath] relation
[imath]2^1 = 2[/imath] relations
[imath]2^3 = 8[/imath] relations
[imath]0 = \binom{1}{2}[/imath]
[imath]1 = \binom{2}{2}[/imath]
[imath]3 = \binom{3}{2}[/imath]
Do you see the general formula for all relations?
[imath]2^\binom{n}{2}[/imath]
Your restriction is you want to pick exactly [imath](1013)² + (1013) - 2[/imath] unordered pairs out of all possible [imath]\binom{2026}{2}[/imath] pairs.
The answer is [imath]\binom{\binom{2026}{2}}{(1013)² + (1013) - 2}[/imath].

 

[Krishang]

You will run out of fuel if you start with a set that contain 2026 elements. You should start low and look for a pattern. This is what I do.
Also for clarity you should start with definitions.
Reflexive Relation
A relation [imath]R[/imath] on a set [imath]A[/imath] is reflexive if [imath](a,a) \in R [/imath] for every [imath]a \in A[/imath].
Symmetric Relation
A relation [imath]R[/imath] on a set [imath]A[/imath] is symmetric if [imath](a,b) \in R [/imath], then [imath](b,a) \in R [/imath] for all [imath]a,b \in A[/imath].
Read the definitions carefully. If you understand them, you can start with a set with one element only.
[imath]A = \{1\}[/imath]
[imath]R = \{(1,1)\}[/imath]
One relation.
If [imath]A = \{1,2\}[/imath]
[imath]R_1 = \{(1,1),(2,2)\}[/imath]
[imath]R_2 = \{(1,1),(2,2),(1,2),(2,1)\}[/imath]
Two relations.
If [imath]A = \{1,2,3\}[/imath]
[imath]R_1 = \{(1,1),(2,2),(3,3)\}[/imath]
[imath]R_2 = \{(1,1),(2,2),(3,3),(1,2),(2,1)\}[/imath]
[imath]R_3 = \{(1,1),(2,2),(3,3),(1,3),(3,1)\}[/imath]
[imath]R_4 = \{(1,1),(2,2),(3,3),(2,3),(3,2)\}[/imath]
[imath]R_5 = \{(1,1),(2,2),(3,3),(1,2),(2,1),(1,3),(3,1)\}[/imath]
[imath]R_6 = \{(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)\}[/imath]
[imath]R_7 = \{(1,1),(2,2),(3,3),(2,3),(3,2),(1,3),(3,1)\}[/imath]
[imath]R_8 = \{(1,1),(2,2),(3,3),(1,2),(2,1),(1,3),(3,1),(2,3),(3,2)\}[/imath]
Eight relations.
Do you see the pattern?
[imath]2^0 = 1[/imath] relation
[imath]2^1 = 2[/imath] relations
[imath]2^3 = 8[/imath] relations
[imath]0 = \binom{1}{2}[/imath]
[imath]1 = \binom{2}{2}[/imath]
[imath]3 = \binom{3}{2}[/imath]
Do you see the general formula for all relations?
[imath]2^\binom{n}{2}[/imath]
Your restriction is you want to pick exactly [imath](1013)² + (1013) - 2[/imath] unordered pairs out of all possible [imath]\binom{2026}{2}[/imath] pairs.
The answer is [imath]\binom{\binom{2026}{2}}{(1013)² + (1013) - 2}[/imath].

Thank you so much for the help