Forming Numbers from Digits

Fill positions left to right. Distinct digits: first slot has (non-zero choices), each later slot one fewer. Leading-zero rule: if 0 is available, the first digit has one fewer option; equivalently total arrangements minus those starting with 0. Repetition allowed multiplies the per-slot choices.

• ÷2: units even. ÷5: units 0 or 5. ÷10: units 0.
• ÷4: last two digits form a multiple of 4. ÷8: last three.
• ÷3 / ÷9: digit sum divisible by 3 / 9 (independent of order — so a fixed digit set is all-or-nothing).
• ÷6: divisible by 2 and 3 together.
• Counting the multiples of $d$ in a RANGE $[a,b]$: $\left\lfloor\tfrac bd\right\rfloor-\left\lfloor\tfrac{a-1}{d}\right\rfloor$, or $\dfrac{\text{last}-\text{first}}{d}+1$ once the first and last multiples are found — 3-digit multiples of 7 run from 105 to 994: $\tfrac{994-105}{7}+1=128$.

Using $n$ distinct digits to form all $n$-digit numbers: each digit appears in each place $(n-1)!$ times. Sum $=(n-1)!\times(\text{sum of digits})\times\underbrace{111\ldots1}_{n}$. Adjust the repeat count and place-value string for $r$-digit selections. When 0 is one of the digits the symmetry breaks in the LEADING place: 0 cannot lead, so each non-zero digit appears in the leading place $(n-1)!$ times but in every other place only $(n-1)!-(n-2)!$ times (the arrangements with 0 leading are removed). Compute the leading-place total and the other places separately, or take the all-digits sum above and subtract the sum of the $(n-1)$-digit numbers formed when 0 leads.