Animated Solution for Mathematics - Permutations and Combinations: If all the six digit numbers $x_1{x}_2{x}_3{x}_4{x}_5{x}_6$ with $0<x_1<x_2<x_3<x_4<x_5<x_6$ are arranged in the increasing order, then the sum of the digits in the $72^{th}$ number is ____.

Imagine you are standing at the threshold of a vast, silent library. In front of you lies a sequence of six-digit numbers, all strictly increasing, all waiting to be counted. The problem asks us to find the $72^{nd}$ number in this sequence.

The constraint $0<x_1<x_2<x_3<x_4<x_5<x_6$ is not a burden; it is a gift. Because the digits must be strictly increasing, the moment you select a set of six distinct digits from the set $\{1,2,3,4,5,6,7,8,9\}$, their order is already decided.

There is only one way to arrange them to satisfy the condition. Therefore, we are not dealing with permutations; we are dealing with combinations. Selection is arrangement.

To find the $72^{nd}$ number, we must act like a librarian organizing books. We need to count systematically. We start with the smallest possible numbers, those that begin with $x_1=1$.

If we fix $x_1=1$, we have five empty slots remaining. We must fill these slots using the digits from the set $\{2,3,4,5,6,7,8,9\}$. This is a set of $8$ available digits. The number of ways to choose $5$ digits from these $8$ is given by the combination formula:

So, there are exactly $56$ numbers that start with $1$. We are still short of our target of $72$, so we must continue our search.

Since $56<72$, we know the $72^{nd}$ number must start with a digit greater than $1$. Let us move to numbers starting with $x_1=2$. To be systematic, we fix the first two digits as $2$ and $3$.

Now, we have $4$ slots left to fill, and we must choose from the digits $\{4,5,6,7,8,9\}$. There are $6$ such digits. The number of ways to choose $4$ digits from these $6$ is:

Now, let us add this to our previous count. We had $56$ numbers starting with $1$. Adding the $15$ numbers that start with $2$ and $3$, we get $56+15=71$.

We are incredibly close! The $71^{st}$ number is the very last number that starts with $2$ and $3$. This means the $72^{nd}$ number must be the very first number that starts with $2$ and $4$.

To find the $72^{nd}$ number, we need the smallest possible number that starts with $2$ and $4$. This means we must pick the smallest available digits for the remaining four slots.

The available digits are $\{5,6,7,8,9\}$. To keep the number as small as possible, we choose the smallest four: $5,6,7,$ and $8$. Thus, our $72^{nd}$ number is $245678$.

Finally, the question asks for the sum of the digits of this number. We simply add them up: $2+4+5+6+7+8$.

The sum is $32$. It is a beautiful, elegant conclusion to a systematic journey. Remember, in JEE Advanced, it is rarely about brute force; it is about finding the structure within the chaos.