The correct option is C 135104
Total cases =9(104)
First case: Choose two non-zero digits = 9C2
Now, number of 5-digit numbers containing both digits =25−2.
Second case: Choose one non-zero and one zero as digit = 9C1
Number of 5-digit numbers containing one non-zero and one zero both =24−1.
Favourable cases = 9C2(25−2)+ 9C1(24−1)
=1080+135=1215
Required Probability =12159×104=135104