How many 6 digit numbers can be formed from two sets of digit? There are two sets of digit :
$ \text{set 1 :} \{~0,1,2,3,4~\}$
$ \text{set 2 :} \{~5,6,7,8,9~\}$
Now how many 6 digit number can we make by taking numbers from these two sets ? From $\text{set 1}$ repetition is permitted but from $\text{set 2}$ repetition is prohibited .
Solution :
The possible combinations are :
\begin{array}{c|cccc}
\ combo &\text{group 1}&  \text{group 2} & \text{without considering 0} &\text{considering 0}&\text{result}\\
\hline
1 & 6 & \fbox0 & 5^6 & 5^6 - 5^5 & 12,500\\
2 & 5 & \fbox1 & 5 \times 5 \times 5 \times 5 \times 5 \times \fbox 5& 5 \times 5 \times 5 \times 5 \times 5 \times \fbox 5  - 5^5 & 12,500\\
3 & 4 & \fbox2 & 5 \times 5 \times 5 \times 5 \times \fbox {$5 \times  4 $}& (5 \times 5 \times 5 \times 5 \times \fbox {$5 \times  4 $})- (5 \times 5 \times 5 \times \fbox {$5 \times  4 $}) & 10,000\\
4 & 3 & \fbox3 & 5 \times 5 \times 5 \times \fbox {$5 \times  4 \times  3 $}& (5\times 5 \times 5 \times \fbox {$5 \times  4 \times  3 $})-(5 \times 5 \times \fbox{$5 \times  4 \times  3 $}) & 6000 \\
5 & 2 & \fbox4 & 5 \times 5 \times \fbox {$5 \times  4 \times  3 \times  2$} & (5 \times 5 \times \fbox {$5 \times  4 \times  3 \times  2$}) - ( 5 \times \fbox {$5 \times  4 \times  3 \times  2$}) & 2400\\
6 & 1 & \fbox5 & 5 \times \fbox {$5 \times 4 \times  3 \times  2 \times  1$} & (5 \times \fbox {$5 \times 4 \times  3 \times  2 \times  1$}) - (\fbox {$5 \times 4 \times  3 \times  2 \times  1$}) & 480\\
\end{array}
So,total number of possible combinations is $= 43880$
Is this procedure correct ?
 A: Yes, but I don't see you taking into account the positioning of the digits. For example, in combo 2, the digit from group 2 can be in one of the 6 places, while you let it be only in the last (it will affect also the 0 consideration).
Sanity check: assume we don't allow the usage of any digit twice. There are 9∗9∗7∗6∗5∗4=68K possible numbers like that. If you allow using some of the digits twice, you should get more options. Your answer, 43K is too small
Solution:
You can write an explicit formula in the following way: $k$ out of the $6$ digits should be from group 1. We need to choose where they will be placed - ${6 \choose k}$, and in each location choose a digit - $5^k$.
For the rest $6-k$ locations - you need to choose which digits will be placed there and in what order: ${5 \choose 6-k}(6-k)!=\tfrac{5!}{(k-1)!}$.
Now, sum over all possi$k$ to obtain
$$\sum\limits_{k=1}^6 {6 \choose k} 5^k  \tfrac{5!}{(k-1)!}=495475$$
From that, we need to remove the ones that start with a $0$ (he is chosen to be at place 1):
$$\sum\limits_{k=1}^6 {5 \choose k-1} 5^{k-1}  \tfrac{5!}{(k-1)!}=61870$$
Total: 433605.
