We Have Liftoff!

To create the largest such number, we are looking for the largest digit $a$ and smallest digit $k$ such that the sequence of seven distinct digits in the desired number is of the form

$a .... a - k .... a - 1 .... a - k - 1 .... a - 2 ..... a - k - 2 ..... a - 3.$

This is designed to make "best use" of the larger digits while still maintaining the "countdown" condition. Essentially we have intertwined two descending sequences, namely $\{a, a-1 , a-2, a-3\}$ and $\{a - k, a - k - 1, a - k - 2\}$ .

Now for all the digits to be distinct, we will require that $a - k \lt a - 3 \Longrightarrow k \gt 3.$ The least such integer $k$ is then $4,$ and taking $a = 9$ in the optimal case we form the number

$9 .... 9 - 4 .... 8 ... 9 - 5 .... 7 .... 9 - 6 ... 6 \Longrightarrow \large\boxed{9584736}.$

As there are $10$ distinct digits, we can use this approach to find the largest "Distinct-digit Countdown number" $C_{n}$ for $1 \le n \le 10,$ (with the appropriate variation for even $n$ .) We then have that

$C_{1} = 9, C_{2} = 98, C_{3} = 978, C_{4} = 9786, C_{5} = 96857, C_{6} = 968574, C_{7} = 9584736,$

$C_{8} = 95847362, C_{9} = 948372615, C_{10} = 9483726150.$