Problem Solving

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

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]