Animated Solution for Mathematics - Permutations and Combinations: If number of 5 digit numbers which can be formed without repeating any digit while tenth place of all of the numbers must be 2 is 336 k find value of k

Imagine you are an architect, but instead of steel and glass, you are building with the very fabric of numbers. You have been tasked with constructing a $5$-digit number, but there are rules—strict, unyielding constraints that define the structure of your creation.

This is the essence of combinatorics: it is not just about counting; it is about understanding the geometry of possibilities.

To begin, visualize five empty pedestals standing in a row, representing the ten-thousands, thousands, hundreds, tens, and units places. We have a pool of ten digits: $\{0,1,2,3,4,5,6,7,8,9\}$.

The golden rule here is that no digit can be repeated. Once a digit is placed on a pedestal, it is removed from our inventory. This is a game of diminishing returns, where every choice you make narrows the path for the next.

We start with the most rigid constraint. The problem demands that the tens place must be the digit $2$.

This is our anchor. Because this position is locked, we have no freedom here—there is only $1$ way to fill this slot. We place the $2$ and move on, knowing that this digit is now permanently unavailable for any other position.

Now, we turn to the most critical position: the ten-thousands place. This is the gatekeeper.

If we place a $0$ here, the entire structure collapses; a number starting with $0$ is not a $5$-digit number, but a $4$-digit one in disguise. Furthermore, we cannot use the $2$ because it is already guarding the tens place.

So, out of our original $10$ digits, we must exclude both $0$ and $2$. This leaves us with $10-2=8$ possible candidates for this first pedestal.

Next, we move to the thousands place. Here, the tension eases.

We have used one digit for the ten-thousands place and the digit $2$ for the tens place. That is two digits gone from our pool. But wait—the digit $0$ is now allowed to return!

It can safely reside in the thousands place without violating any rules. Therefore, we have $10$ total digits minus the $2$ already used, giving us $8$ choices for this position as well.

As we move to the hundreds place, the pattern becomes clear. We have already occupied three positions: the ten-thousands, the thousands, and the tens.

With three distinct digits removed from our pool of ten, we are left with $10-3=7$ choices. Finally, for the units place, we have filled four positions, leaving us with $10-4=6$ choices.

By the Fundamental Principle of Counting, the total number of valid configurations is the product of the choices for each independent position:

Let us calculate this carefully. $8\times 8=64$. Then, $64\times 7=448$. Finally, $448\times 6=2688$.

We have successfully constructed $2688$ unique $5$-digit numbers that satisfy all our constraints.

The problem states that this total is equal to $336k$. We have done the heavy lifting; now we simply equate our result to the given expression:

To find $k$, we divide $2688$ by $336$. A quick calculation reveals that $k=8$.

And there you have it. By systematically respecting the constraints and carefully managing our pool of digits, we have navigated the complexity of the problem. Remember, in combinatorics, the secret is always to tackle the most restricted positions first.