Here, p,pn∈1,2,…50
Now p can take values
2,3,5,7,11,13,17,23,29,31,37,41,43 and 47.
∴ R1={(2,{2}^{0}),(2,{2}^{1})\ldots (2,{2}^{5}),(3,30),…(3,33),(5,50),…(5,52),(7,70),…(7,72),(11,{11}^{0}),(11,{11}^{1}),...(47,{47}^{0}),(47,{47}^{1})}
R2=(2,20),(2,21),(3,30),(3,31)....(47,470),(47,471)
Now R1−R2≡R1∩R2′
So R1−R2=(2,22),(2,23),(2,24),(2,25),(3,32),(3,33),(5,52),(7,72)
∴n(R1−R2)=8