Animated Solution for Mathematics - Permutations and Combinations: The total number of three-digit numbers, with one digit repeated exactly two times, is ______.

Mastering the Zero Trap

Welcome, future engineer! Today, we are not just solving a combinatorics problem; we are learning to think like a mathematician. When you face a problem like 'counting 3-digit numbers with exactly one digit repeated twice,' your first instinct might be to jump straight into a formula.

Resist that urge! The beauty of JEE Advanced problems lies in their ability to punish those who rush and reward those who visualize.

The Rebel Digit

In the world of 3-digit numbers, the digit zero is a rebel. It refuses to sit in the hundreds place.

If we ignore this, we fall into the trap of counting numbers like $055$, which is mathematically just $55$—a 2-digit number. To conquer this, we must partition our universe into three mutually exclusive cases based on the presence of zero.

The Zero-Heavy Cases

Let's tackle the cases where zero is present.

Here, the number must look like $d00$, where $d$ is a non-zero digit. Since $d$ can be any value from $\{1,2,\dots ,9\}$, we have exactly $9$ choices.

This is our first batch of numbers: $9$.

Here, some other non-zero digit $d$ must be the one that repeats. The zero cannot be at the front, so our valid forms are $dd0$ and $d0d$.

We have $9$ choices for $d$, and for each $d$, we have $2$ possible arrangements. Thus, we have $9\times 2=18$ possibilities.

The Grand Finale (No Zeros)

Now, we enter the most expansive territory: Case 3, where zero is not used at all. All our digits must come from the set $\{1,2,\dots ,9\}$.

We need to form a 3-digit number with exactly one digit repeated twice. This means we need two distinct digits, say $a$ and $b$.

First, we select these two digits from the nine available:

Next, we must decide which of these two digits is the one that repeats. We have $2$ choices for the repeating digit.

Finally, we arrange these three digits (e.g., $a,a,b$). The number of permutations of three items where two are identical is given by:

So, the total for Case 3 is $36\times 2\times 3=216$.

We have systematically dismantled the problem. By summing our cases, we arrive at the final count:

Remember, the math is not just about the final number; it is about the logic that gets you there. You have successfully navigated the constraints, avoided the zero-trap, and executed the combinatorics with precision.

Keep this structured approach in your toolkit, and no problem will ever be too complex for you! The final answer is 243.