Write a 9-digit number where exchanging any two digits results in a bigger number. How many such numbers exist?

Question

Homi · Accepted Answer

The condition means the digits must be in strictly increasing order.

Step 1 — Understanding the condition

Let us call the 9-digit number $N$ . Let its digits be $d_1 d_2 d_3 d_4 d_5 d_6 d_7 d_8 d_9$ . $d_1$ is the first digit. $d_9$ is the last digit. We swap any two digits. Let us swap $d_i$ and $d_j$ . Assume $i$ is smaller than $j$ . This means $d_i$ is to the left of $d_j$ . The new number is $N'$ . We want $N'$ to be bigger than $N$ . When comparing numbers, we look from left to right. The first different digit decides which number is larger. If we swap $d_i$ and $d_j$ . Digits before $d_i$ stay the same. Digits after $d_j$ also stay the same. The new number $N'$ has $d_j$ at position $i$ . The old number $N$ has $d_i$ at position $i$ . For $N'$ to be larger, $d_j$ must be greater than $d_i$ . So, $d_i < d_j$ must be true. This must be true for any pair of digits. This means the digits must be in strictly increasing order. So, $d_1 < d_2 < d_3 < d_4 < d_5 < d_6 < d_7 < d_8 < d_9$ .

Step 2 — Finding one such number

The digits must be distinct. They must be in increasing order. A 9-digit number cannot start with zero. So, $d_1$ cannot be $\mathbf{0}$ . The smallest possible digit for $d_1$ is $\mathbf{1}$ . Since $d_1 < d_2 < ... < d_9$ , all digits must be different. We need 9 distinct digits. The available digits are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Since $d_1$ must be at least 1. The smallest possible digits are 1, 2, 3, 4, 5, 6, 7, 8, 9. Arranging these in increasing order gives the number. The first digit is $\mathbf{1}$ . The second digit is $\mathbf{2}$ . The ninth digit is $\mathbf{9}$ . The number is 123456789.

Let us check this number. Original number is 123456789. Swap 1 and 2: 213456789. This is bigger. Swap 3 and 7: 127456389. This is bigger. This number works.

Diagram 1

Step 3 — Counting how many such numbers exist

We need to choose 9 distinct digits. These digits must be from 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Once chosen, they must be arranged in increasing order. This is the only way to satisfy the condition. The first digit $d_1$ cannot be $\mathbf{0}$ . If we choose 9 digits that include $\mathbf{0}$ . Then $\mathbf{0}$ will be $d_1$ . This would not be a 9-digit number. So, we must choose 9 digits that do not include $\mathbf{0}$ . The available digits without $\mathbf{0}$ are 1, 2, 3, 4, 5, 6, 7, 8, 9. There are $\mathbf{9}$ such digits. We need to choose $\mathbf{9}$ digits from these $\mathbf{9}$ digits. There is only one way to choose them. We must choose all of them. The chosen digits are {1, 2, 3, 4, 5, 6, 7, 8, 9}. Arranging them in increasing order gives 123456789. This is the only number that fits all conditions. So, there is only one such number.

Answer

(i) A 9-digit number where exchanging any two digits results in a bigger number is 123456789. (ii) There is only 1 such number.

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