If we take out p common from second bracket, function will be 0 when p=0

Take cases of p=1

Now when p=1 and because of third bracket when Q =1 function will be 0.

Hence we are left only with two min terms (P,Q,R) = 100 + 101.

Try doing such examples using truth table in exam won't take much time and increases accuracy.

Take cases of p=1

Now when p=1 and because of third bracket when Q =1 function will be 0.

Hence we are left only with two min terms (P,Q,R) = 100 + 101.

Try doing such examples using truth table in exam won't take much time and increases accuracy.