find amount all the 3-digit numbers The problem states:
find an amount of 3-digit numbers with digits that are not the power of cube (as far as I understand $8$ can not be taken as it is $2^3$ I am also not sure about $1^3 = 1$) and also these numbers have to have digits that go in descending order. If take the combinatory formula it is gonna be "number of combinations w/o repeation" (because I also have to write final calсulus using the formula).
it is also clear that there are 3 sequences of digits


*

*First position - 6 digits ( I excluded $0$, $1$ and $8$)

*Second position - 7 digits (I excluded $1$ and $8$)

*Thrid position - 7 digits (I excluded $1$ and $8$)


still not sure how to arrange "in descending order" part
 A: Without repetition, the answer is $\binom 73=35$.
This is because the order is automatically determined when you choose three numbers from $2,3,4,5,6,7$ or $9$.
A: First, let $abc$ be a $3-$digit number where $a \ne 0,1,8$, $b \ne 0,1,8$ and $c \ne 0,1,8$.
Now, notice that since $a \ne 0,1,8$ and $b \ne 0,1,8$, there are $7 \cdot 7 = 49$ of $2$-digit numbers $ab$. Now assuming equality is not included in descending order, we can eliminate the cases where $a=b$ and there are $7$ of them (namely $22$, $33$, $44$, $55$, $66$, $77$, $99$). And among $49-7=42$ numbers, in half of them $a < b$ and in half of them $b < a$ (we can use symmetry because $0$ is already eliminated so all of those $42$ numbers are valid $2$-digit numbers). So there are $21$ numbers here only with the condition $a > b$.
By using the same argument, we can say that there are $21$ numbers only with condition $b > c$. However, in this case, we are also counting the cases where $b = 9$ and there are $6$ of them so we have to exclude them. Also notice that we have to exclude the case where $a = 9$ and $b = 2$ because $c \ne 0,1$. So there are $7$ cases to exclude and the answer becomes $21+21-7 = 35$.
However, I don't know if an argument like this one works in for example $4$-digit numbers. And this is not a "formal" way neither.
A: Another more explicit way than directly giving combinations.
I take strict descending order and digits $9,7,6,5,4,3,2$ ($0,1,8$ are assumed to be cubes). we have $abc$ with $ a\gt b\gt c$
$$\begin{array}aa=9\Rightarrow 5+4+3+2+1\text{ possible numbers }\\a=7\Rightarrow 4+3+2+1\text{ possible numbers }\\a=6\Rightarrow 3+2+1\text{ possible numbers }\\a=5\Rightarrow 2+1\text{ possible numbers }\\a=4\Rightarrow 1\text{ possible number: 432  }\\a=3\Rightarrow\text{no possible numbers }\end{array}$$
Thus there are $15+10+6+3+1=35$ numbers
