Question:

Step 1: Understand the nature of the numbers being counted. The numerical results $70$ and $35$ in the solution, along with the subsequent identification of the $105^{th}$ number, suggest that the problem involves counting five-digit numbers with strictly increasing digits. A five-digit number $d_1 d_2 d_3 d_4 d_5$ with strictly increasing digits implies $0 < d_1 < d_2 < d_3 < d_4 < d_5 \le 9$. This means all five digits must be distinct and chosen from the set $\{1, 2, \dots, 9\}$. Once five distinct digits are chosen, there is only one way to arrange them in strictly increasing order to form such a number. Step 2: Calculate the count of five-digit numbers with strictly increasing digits that start with 1. For a number to start with 1 ($d_1 = 1$), the remaining four digits ($d_2, d_3, d_4, d_5$) must be chosen from the digits strictly greater than 1. These digits are from the set $\{2, 3, 4, 5, 6, 7, 8, 9\}$, which contains 8 distinct digits. The number of ways to choose 4 distinct digits from these 8 digits is given by the combination formula $\binom{n}{k}$. $$ \text{Number of such numbers} = \binom{8}{4} $$ $$ \binom{8}{4} = \frac{8!}{4!(8-4)!} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70 $$ These are the first 70 such numbers when ordered sequentially (e.g., from $12345$ to $16789$). Step 3: Calculate the count of five-digit numbers with strictly increasing digits that start with 2. Following the numbers that start with 1, we now consider numbers starting with 2 ($d_1 = 2$). For these numbers, the remaining four digits ($d_2, d_3, d_4, d_5$) must be chosen from the digits strictly greater than 2. These digits are from the set $\{3, 4, 5, 6, 7, 8, 9\}$, which contains 7 distinct digits. The number of ways to choose 4 distinct digits from these 7 digits is given by the combination formula $\binom{n}{k}$. $$ \text{Number of such numbers} = \binom{7}{4} $$ $$ \binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 $$ These 35 numbers immediately follow the 70 numbers that start with 1. Step 4: Calculate the total count of numbers up to those starting with 2. The total number of five-digit numbers with strictly increasing digits that start with 1 or 2 is the sum of the counts from Step 2 and Step 3. $$ \text{Total numbers} = (\text{Numbers starting with 1}) + (\text{Numbers starting with 2}) $$ $$ \text{Total numbers} = 70 + 35 = 105 $$ This means the $105^{th}$ number in this ordered sequence is the last number among those that start with 2. Step 5: Identify the $105^{th}$ five-digit number with strictly increasing digits. The $105^{th}$ number is the largest five-digit number with strictly increasing digits that starts with 2. To form the largest such number, we fix the first digit as 2, and then choose the largest possible remaining four digits from the available set $\{3, 4, 5, 6, 7, 8, 9\}$. The largest four digits in this set are $\{6, 7, 8, 9\}$. Arranging these digits in increasing order along with the starting digit 2 gives the number: $$ 26789 $$ The final answer, as derived from the provided solution's logic for a combinatorics problem, is $26789$.