For the resulting number to be a multiple of 30, it must be a multiple of 3 and 10. To be divisible by 10, the last digit must be 0 Hence d is 0. Note that we can't substitute the digit d otherwise the units digit will be the digit 1. One of the numbers from $\{a, b, c\}$ will be 1 so the remaining two numbers $k$ and $m$ must satisfy $k+m \equiv 2 (\mod 3)$. The possible remaining $k$ and $m$ values are: (8, 6), (8, 3), (7, 4), (7, 1), (6, 5), (6, 2), (5, 3), (4, 1), (3, 2). There are then a total of 27 unordered sets of the remaining digits. where the value substituted has to be greater than $k$ and $m$. Furthermore, there are $3!$ ways of arranging these digits. Hence, the answer is $6 \cdot 27 = \boxed{162}$

Without restrictions, there are a total of $9\cdot 9 \cdot 8 \cdot 7=4536$ combinations. In order for the number to be a multiple of $30$, it must be divisible by $10$ and $3$, in which the restriction that it is divisible by $10$ will make $d$ be equal to $0$. Fixing $0$ to $d$ make the total combinations equal to $9\cdot 8 \cdot 7 \cdot 1=504$ combinations. Note that one number must be $1$.Now, we must restrict the number to be a multiple of $3$. The divisibility rule of $3$ states that the sum of the digits must be divisible by $3$ in order to be divisible such that the remaining two digits $b$ and $c$ satisfy the following: $b+c=5$, $b+c=8$, $b+c=11$, $b+c=14$, and $b+c=17$ are the possible sums. The values of $b$ and $c$ will be in the ordered pairs of $(3,2)$, $(4,1)$, $(5,3)$, $(6,2)$, $(6,5)$, $(7,1)$, $(7,4)$, $(8,3)$, and $(8,6)$. I believe there are a total of $27$ sets of the remaining digits in which the Replaced value, $x$, is $x>b$ and $x>c$. Also, there a total of $6$ ways to arrange the digits; Hence the answer is $27 \cdot 6$, or $162$ numbers that satisfy the conditions.

This guy is sitting right next to me btw ^^^

This guy is actually telling the truth ^^^

OofPirate wrote: This guy is actually telling the truth ^^^ r u 2 brothers or r u 2 at skewl and ur sitting nekst 2 eetch uhther?

We are at the skewl