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 \(x \neq 0\) ?

• A. 245
• B. 285
• C. 240
• D. 320

Hint:

Fix the middle digit y first. Count how many hundreds digits and how many units digits are allowed for that y, then add over all y.

Answer 1

The Correct Option is C

Step 1: Understand the conditions:
The number is \(xyz\) with \(x\) in the hundred's place. The conditions say that the middle digit y is larger than both the first digit x and the last digit z. Also \(x \neq 0\) so that it is a true three digit number.
Digits run from 0 to 9. Since x is at least 1 and x must be less than y, y is at least 2.

Step 2: Fix y and count the choices:
For a given y, the digit x can be any of \(1, 2, \ldots, y-1\). That is \(y-1\) choices (x cannot be 0).
The digit z can be any of \(0, 1, \ldots, y-1\). That is \(y\) choices (z may be 0).
The choices of x and z are independent, so for a given y there are \(y(y-1)\) numbers.

Step 3: Add over every y:
\[ \sum_{y=1}^{9} y(y-1) = \sum_{y=1}^{9} y^2 - \sum_{y=1}^{9} y \]
We know \(\sum y^2 = \tfrac{9 \cdot 10 \cdot 19}{6} = 285\) and \(\sum y = \tfrac{9 \cdot 10}{2} = 45\).
\[ 285 - 45 = 240 \]
Listing the terms as a check: 0 + 2 + 6 + 12 + 20 + 30 + 42 + 56 + 72 = 240.

Step 4: Why the other options are wrong:
The number 285 is what we would get if x were allowed to be 0 as well, because then x would have y choices. It wrongly includes 45 numbers that start with 0, which are not three digit numbers. The values 245 and 320 do not come from any consistent counting. Only 240 matches the careful sum.

Final Answer:
There are 240 such numbers. \[ \boxed{\text{Option (3): } 240} \]