Question

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 $x\ne y$, z < y and $x\ne 0$?

• A. $245$
• B. $285$
• C. $240$
• D. $320$

Answer 1

Answer: (C)

$240$

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=$1\times 2$.
For y=3, x=1,2 and z = 0,1,2 total numbers possible=$2\times 3$.
For y=4, x=1, 2, 3 and z = 0,1,2,3 total numbers possible=$3\times 4$.

In the same manner, For y=9, total numbers possible = $8\times 9$.
Hence, the total numbers of three digit positive integers is =$1\times 2+2\times 3+3\times 4+4\times 5+5\times 6+6\times 7+7\times 8+8\times 9=240$.