How many three digit positive integers, with digits x, y and z in the hundred's, ten's and unit's place respectively, exist such that x <y, z< y and x=!0?
a. 245
b. 285
c. 240
d. 320
x, y and z in the hundred's, ten's and unit's place.
So y should start from 2
If y=2 , possible values of x=1 and z = 0,1 .
So 2 cases 120,121.
Also if y=3 , possible values of x=1,2 and z=0,1,2.
Here 6 three digit nos. possible .
Similarly for next cases would be 3*4=12,4*5=20,5*6=30,.....,8*9=72 .
Adding all we get 240 cases...Write Here