It is evident that middle digit of the number is to be the largest of three digits in the number as per condition.
For y=1,no numbers possible as x is not equal to zero.
For y=2,x=1 and z=0,1 total numbers possible=1*2
For y=3,x=1,2 and z=0,1,2 total numbers possible=2*3
For y=4,x=1,2,3 and z=0,1,2,3 total numbers possible=3*4
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For y=9,total numbers possible=8*9
Total numbers=1*2+2*3+3*4+.........+8*9=240 Option A
P.S.We can get the sum of above series by writing the general term as n(n+1)=n^2+n and using the formulas for sum of squares of natural numbers and sum of natural numbers.

sanju09 wrote: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) 240
(B) 245
(C) 285
(D) 320
(E) 720

The middle digit y can take values from 2-9

when y is 9 x can take values from 1-8 and z can take values from 0-8
Total num = 8*9
when y is 8 x can take values from 1-7 and z can take values from 0-7
Total num = 7*8

....

[spoiler]Total num = 9*8+8*7+7*6+6*5+5*4+4*3+3*2+2*1 =240[/spoiler]