281 views

1 Answer

0 votes
0 votes

a) and b)

No of relations in both (a) and (b) = $\begin{align*} 2^{\left ( n^2 - 1 \right )} \end{align*}$

(c) 

No of relations such that no pair in $R$ has $a$ as its first element = $\begin{align*} 2^{\left ( n^2 - n \right )} \end{align*}$

(d)

No of relations such that at least one ordered pair in $R$ has $a$ as its first element = $\begin{align*} 2^{\left ( n^2 \right )} - 2^{\left ( n^2 - n \right )} \end{align*}$

(e)

Here we are not allowing a as first element or b as second element of any pair ($x$,$y$) $\in$ $R$

No of such relation = $\begin{align*} 2^{n^2 - \left ( 2n-1 \right )} \end{align*}$

(f)

Here we must have atleast one pair ($x$,$y$) such that $a$ is the first element or $b$ is second element.

No of such relation = $\begin{align*} 2^{n^2} - 2^{n^2 - \left ( 2n-1 \right )} \end{align*}$

Related questions

0 votes
0 votes
1 answer
2
srestha asked May 15, 2018
1,150 views
How to distinguish between countably finite , countably infinite , uncountably infinite set?for reference see this ques:https://gateoverflow.in/36654/why-set-of-all-funct...
2 votes
2 votes
1 answer
4
ram_18051996 asked Jun 15, 2017
516 views
{ a } ∈ A buta ∉ Awhy ?here ' a is the element of set {a} ' ,and ' set {a} is the element of A" , so " a also element of A " . please clear my doubt .