Question 81

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 $$\neq$$ 0 ?

Solution

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.

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