number of possibilities to interview candidates Given 8 candidates for a job, Calculate the possibilities to interview them in one day if each one is interviewed at least once and no more than 3 times.
Note: the total number of interviews is 13 and that order matters.
Here is my suggested answer which I was told is wrong:

could someone help me find the correct one
 A: At most 2 candidates are interviwed 3 times.


*

*Exactly 2 are. Then exactly one is twice and other once. First Chose 2 from that 8 and then 1 from 6 and then arrange them all in line, so you have $${8\choose 2} \cdot {6\choose 1} \cdot {13!\over 3!3!2!}=13!{7\over 3}$$

*Exactly 1 is. Then exactly three are twice and four once. First Chose 1 from that 8 and then 3 from 7 and then arrange them all in line, so you have $${8\choose 1} \cdot {7\choose 3} \cdot {13!\over 3!2!^3}=13!{35\over 6}$$

*No one is 3 times . Then exactly five are twice and three once. First Chose 5 from that 8 and then arrange them all in line, so you have $${8\choose 5} \cdot {13!\over 2!^5}=13!{7\over 4}$$
Sum all that and we get $11!\cdot 91\cdot 17$
A: You can tell that there are 3 categories here to make $A+B+C+D+E+F+G+H=13$:
A   B   C   D   E   F   G   H       SUM
2   2   2   2   2   1   1   1       13
3   3   2   1   1   1   1   1       13
3   2   2   2   1   1   1   1       13

If you look at the first row you can arrange it with: $\frac{8!}{(3!*5!)} = 56$ ways.
Sum all categories together you get: $56+168+280=504$
