Combinatorics – How Many Seven-Digit Numbers Have Seven Distinct Nonzero Digits in Increasing Order?

combinatorics

How many seven-digit numbers have seven distinct nonzero digits that appear in increasing order from left to right?

If we were not restricted by the increasing order of digits, the solution would be $\binom{9}{7}\cdot7!=36\cdot7!=181\, 440$

However, the requirement of increasing order of digits truly throws a wrench into my plans.

I have noticed that the first digit must be less than or equal to $3$, so there are $3$ possible numbers for the first digit.

Then, we can perform casework to determine the number of possible numbers for the second digit.

  1. If the first digit is $3$, then the second digit must be $4$ (1 option).
  2. If the first digit is $2$, then the second digit can either be $3$ or $4$ (2 options).
  3. If the first digit is $1$, then the second can either be $2$, $3$, or $4$ (3 options).

So, there are $6$ possible second digits.

However, continuing down this path is highly tedious and prone to error. Is there a better way of solving this problem?

Best Answer

For any seven distinct numbers you chose out of the possible nine, there is only one way to arrange them in an increasing order. So, the answer is just $\binom97$ without the $7!$.

Hope this helps. :)

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