propositional logic formulas is TRUE when exactly two of p,q and r are TRUE means possibilities are pq or qr or pr to be TRUE

**option A)**

((p↔q)∧r)∨(p∧q∧∼r)

=((T↔T)∧r)∨(T∧T∧∼r) ( take p=T and q=T)

=((T)∧r)∨(T∧∼r)

=(r)∨(∼r)

=T

It doesn't depend upon the value of r, so, if the value of r is T then it also to be T so the condition exactly 2 are T is false. Hence **option A is FALSE.**

**option B)**

(∼(p↔q)∧r)∨(p∧q∧∼r)

=(∼(T↔T)∧r)∨(T∧T∧∼r) ( take p=T and q=T)

=(∼(T)∧r)∨(T∧∼r)

=(F∧r)∨(T∧∼r)

=(F)∨(∼r)

=∼r

so if r=F then this is T so condition exactly 2 are T is true

(∼(p↔q)∧r)∨(p∧q∧∼r)

=(∼(T↔q)∧T)∨(T∧q∧∼T) ( take p=T and r=T)

=(∼(q)∧T)∨(F)

=∼(q)∨(F)

=∼q

so if q=F then this is T so condition exactly 2 are T is true

(∼(p↔q)∧r)∨(p∧q∧∼r)

=(∼(p↔T)∧T)∨(p∧T∧∼T) ( take q=T and r=T)

=(∼(p)∧T)∨(F)

=∼(p)∨(F)

=∼p

so if p=F then this is T so condition exactly 2 are T is true

**Hence option B is CORRECT.**

**option C)**

((p→q)∧r)∨(p∧q∧∼r)

=((T→T)∧r)∨(T∧T∧∼r) ( take p=T and q=T)

=((T)∧r)∨(T∧∼r)

=(r)∨(∼r)

=T

It doesnt depend upon value of r , so , if value of r is T then it also to be T so the condition exactly 2 are T is false . **Hence option C is FALSE.**

**option D)**

(∼(p↔q)∧r)∧(p∧q∧∼r)

=(∼(T↔T)∧r)∧(T∧T∧∼r) ( take q=T and r=T)

=(∼(T)∧r)∧(T∧∼r)

=(F∧r)∧(T∧∼r)

=F∧(∼r)

=F

It doesnt depend upon value of r , so , if value of r is F then it also to be F so the condition exactly 2 are T is false . **Hence option D is FALSE.**