Question:

We need five-digit numbers where each digit strictly exceeds the previous one. This is equivalent to choosing 5 distinct digits from {1,2,3,4,5,6,7,8,9} where 1 cannot be chosen (since we need a five-digit number starting with at least 1, and if we choose 5 digits in increasing order, the first must be at least 1). We're essentially choosing 5 digits from 9 available digits, then arranging them in increasing order—there's only one arrangement. The total count is $\binom{9}{5} = 126$ such numbers. To find the 105th number, we count backwards: numbers NOT containing digit 1 are formed from {2,3,4,5,6,7,8,9}, giving $\binom{8}{5} = 56$ numbers. Numbers NOT containing 2 are from {1,3,4,5,6,7,8,9}, giving $\binom{8}{5} = 56$ numbers. Numbers NOT containing 3 are from {1,2,4,5,6,7,8,9}, giving $\binom{8}{5} = 56$ numbers. Since $56 + 56 + 56 = 168 > 105$, the 105th number must be among the first 56 (not containing 1), first 56 (not containing 2), or first 56 (not containing 3). Checking: the first 56 numbers lack digit 9, the next 56 lack digit 8, and so on. The 105th number = 56 + 49th number without 8 = 49th from {1,2,3,4,5,6,7,9}, which is 12479. This contains digits 1, 2, 4, 7, 9, so it does NOT contain 3 or 5.