How many numbers can be formed from the digits 0, 1, 3, 4, 5, 7, and 9 taken 5 at a time, if every number must contain the digits 0,4,5? 
How many numbers can be formed from the digits 0, 1, 3, 4, 5, 7, and 9 taken 5 at a time, if every number must contain the digits 0,4,5?

For me, I utilized the permutation method, as 5 taken digits can be permuted between them upon choices.
But, every number must contain 0,4,5 so 3 places are preoccupied. 0 can be in 5 places (as the number of the digits of the number to be formed has not been defined in the question), 4 can be in 4 places, 5 can be in 3 places. The remaining two places can be filled by 1, 3, 7, and 9, so $^4P_2$.
Thus, my answer would be 5 * 4 * 3 * 4 * 3. Can somebody pls help me with this problem?
 A: As we know, $0$, $4$ & $5$ have to be in the number formed, so the other two digits have to be chosen from the remaining four options $1$, $3$, $7$ or $9$. Also, we will have to arrange these five digits to get all the possible permutations for this question.
Hence, the answer expression shall be:
$^4C_2 * 5!$
= $6*120$
= $720$
Kindly note that

taken 5 at a time

did not imply that the number has to be a 5 digit number, ie. $0$
can be placed at the first position too! Else, $5!$ should be replaced by $4*4!$
A: We interpret "number" as meaning a string of digits, so the problem asks us to find the number of five-digit strings taken from $\{0,1,3,4,5,7,9\}$ which contain at least one $0$, at least one $4$, and at least one $5$.  It's possible for such a string to start with $0$.  (This is contrary to the usual terminology in combinatorics, where a "number" cannot start with $0$.)
We will use the Principle of Inclusion and Exclusion.  Without the restriction on the required digits, there are $N=7^5$ five-digit strings taken from the given set of digits.
Let's say a five-digit string has "Property $i$" if it does not contain the digit $i$, for $i\in \{0,4,5\}$, and define $S_j$ as the number of strings with $j$ of the properties, for $j = 1,2,3$. To compute $S_j$, note that the $j$ omitted digits can be chosen in $\binom{3}{j}$ ways, and then there are $(7-j)^5$ ways to arrange the remaining $7-j$ digits in a string of length $5$, so
$$S_j = \binom{3}{j}(7-j)^5$$
By inclusion / exclusion, the number of five-digit strings with none of the properties, i.e. the number of strings with at least one each of $0,4$ and $5$,
is
$$N-S_1+S_2-S_3 = \boxed{1830}$$
If we adopt the convention that the number cannot start with $0$ then a solution by inclusion / exclusion is still possible, but it's more complicated.
