Question

The number of 3-digit numbers of the form $xyz$ with $x<y$, $z<y$ and $x\ne0$ is:

• A. 284
• B. 240
• C. 44
• D. 270

Hint:

This series fits the standard summation identity for successive integer products: $\sum_{k=1}^{n} k(k+1) = \frac{n(n+1)(n+2)}{3}$. Substituting $n=8$ (since the final term is $8 \times 9$) gives $\frac{8 \times 9 \times 10}{3} = 240$, bypassing the step-by-step additions entirely!

Answer 1

The Correct Option is B

Concept: This problem can be systematically solved by partitioning the possible outcomes based on the value of the middle peak digit $y$. Since $y$ must be strictly greater than both $x$ and $z$, it acts as the absolute maximum boundary for the individual digits. Step 1: Establish the mathematical constraints for each digit slot.
For a standard three-digit number $xyz$:
• The leading hundred's digit $x$ cannot be zero ($x \in \{1, 2, \dots, 9\}$).
• The middle ten's digit $y$ and unit digit $z$ can be any integer from 0 to 9 ($y, z \in \{0, 1, \dots, 9\}$).
• We are given the structural inequalities: $x < y$ and $z < y$. Because $x \ge 1$ and $y > x$, the minimum possible value for the peak digit $y$ is $2$.

Step 2: Count configurations by looping over the values of $y$.
Let us calculate the independent number of valid choices for $x$ and $z$ for each possible assignment of $y$ from 2 to 9:
• If $y = 2$: - $x$ must satisfy $1 \le x < 2 \implies x \in \{1\}$ (1 choice) - $z$ must satisfy $0 \le z < 2 \implies z \in \{0, 1\}$ (2 choices) - Combinations: $1 \times 2 = 2$
• If $y = 3$: - $x$ must satisfy $1 \le x < 3 \implies x \in \{1, 2\}$ (2 choices) - $z$ must satisfy $0 \le z < 3 \implies z \in \{0, 1, 2\}$ (3 choices) - Combinations: $2 \times 3 = 6$
• If $y = 4$: - Choices: $x \in \{1, 2, 3\}$ (3 choices), $z \in \{0, 1, 2, 3\}$ (4 choices) - Combinations: $3 \times 4 = 12$ Following this structural induction, for any given peak value $y$, the number of choices for $x$ is always $(y - 1)$ and the number of choices for $z$ is always $y$.

Step 3: Sum the total number of valid permutations.
Summing the individual products over the entire range from $y = 2$ to $y = 9$: $$\text{Total Numbers} = \sum_{y=2}^{9} (y - 1)y$$ $$\text{Total} = (1 \times 2) + (2 \times 3) + (3 \times 4) + (4 \times 5) + (5 \times 6) + (6 \times 7) + (7 \times 8) + (8 \times 9)$$ $$\text{Total} = 2 + 6 + 12 + 20 + 30 + 42 + 56 + 72 = 240$$ This aligns perfectly with option (B).