Here, X={1,2,3,4,5}
Now, the possibilities are
(i) 1∈Y,1∈Z
(ii) 1∈Y,1∈/Z
(iii) 1∈/Y,1∈Z
(iv) 1∈/Y,1∈/Z
(ii) (iii) ,(iv) are favourable cases for Y∩Z to be empty.
Similarly , for other elements we have 3 favourable cases. ∴ Required number of different ordered pairs (Y,Z)=3×3×3×3×3=35