Kenneth Rosen Edition 6th Exercise 7.1 Question 44 (Page No. 473)

Question

Let $S$ be a set with $n$ elements and let $a$ and $b$ be distinct elements of $S$. How many relations are there on $S$ such that

(a) $(a,b) \in S$

(b) $(a,b) \not\in S$

(c) There are no ordered pairs in the relation that have "$a$" as their first element.

(d) There is at least one ordered pair in the relation that has "$a$" as its first element.

(e) There are no ordered pairs in the relation that have "$a$" as their first element and there are no ordered pairs in the relation that have "$b$" as their second element.

(f) There is at least one ordered pair in the relation that either has "$a$" at its first element or has $b$ as its second element.

I tried this problem and my answers are
(a) $2^{n^2-1}$
(b) same as (a)

(c) $2^{n(n-1)}$

(d) $2^{n^2}-2^{n(n-1)}$

(e) $2^{(n-1)^2}$

(f) $2^{n^2}-2^{(n-1)^2}$

Please let me know if my work is correct.

Comments

All are correct.
Just have a little confusion in $"f" -$
It allows $(a,b) \in R$ right ?

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yes, It not only allows (a,b)∈R ,but also ( a,___ )∈R and ( ___,b )∈R

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ok, thanks @shaik .

Answer 1

(a) (a,b)∈S
Reasoning: Total relation with n elements on itself = $2^{n^{2}}$     Since (a,b) is present so for (a,b) we have 1 choice, and for remaining ordered pair we have 2 choices . hence Resultant relation = $2^{n^{2}-1}$

(b) (a,b)∉S
Reasoning:Total relation with n elements on itself = $2^{n^{2}}$     Since (a,b) is not present so for (a,b) we have 0 choice, and for remaining ordered pair we have 2 choices . hence Resultant relation = $2^{n^{2}-1}$

(c) There are no ordered pairs in the relation that have "a" as their first element.
Reasoning: Total relation with n elements on itself = $2^{n^{2}}$     Since In (a,b)  a is not present as the first element so we have n-1 choices for choosing a in (a,b)  and n choices for choosing b. So, $2^{n\times (n-1)}$

(d) There is at least one ordered pair in the relation that has "a" as its first element.
Reasoning: Total Relation - question (c) = $2^{n^{2}} - 2^{n\times (n-1)}$

(e) There are no ordered pairs in the relation that have "a" as their first element and there are no ordered pairs in the relation that have "b" as their second element.
Reasoning: Total relation with n elements on itself = $2^{n^{2}}$     Since In (a,b)  a is not present as the first element so we have n-1 choices for choosing a in (a,b)  and n choices for choosing b, similarly for b. So, $2^{ (n-1)\times(n-1)}$

(f) There is at least one ordered pair in the relation that either has "a" at its first element or has b as its second element.
Reasoning: Total relation - question (f) = $2^{ n^{2}} - 2^{ (n-1)\times(n-1)}$