Let the numbers selected from classes 10, 11, 12 be $x,y,z$ respectively.

Then: $x+y+z=10$

Given conditions:

• At least $2$ from each class: $x\ge 2,\ y\ge 2,\ z\ge 2$
• There are only $5,6,8$ students available respectively, so: $x\le 5,\ y\le 6,\ z\le 8$
• At most $5$ students from class 10 and 11 together: $x+y\le 5$

Since $x\ge 2$ and $y\ge 2$, we have: $x+y\ge 4$ Combined with $x+y\le 5$, possible values are only: $x+y=4 \text{ or } 5$

Case 1: $x+y=4$

Since $x,y\ge 2$, the only possibility is: $(x,y)=(2,2)$ Then: $z=10-4=6$ This is valid.

Number of ways: $\binom{5}{2}\binom{6}{2}\binom{8}{6}$ $=10\cdot 15\cdot 28=4200$

Case 2: $x+y=5$

With $x,y\ge 2$, possibilities are: $(x,y)=(2,3),(3,2)$ Then: $z=10-5=5$

For $(x,y,z)=(2,3,5)$: $\binom{5}{2}\binom{6}{3}\binom{8}{5}$ $=10\cdot 20\cdot 56=11200$

For $(x,y,z)=(3,2,5)$: $\binom{5}{3}\binom{6}{2}\binom{8}{5}$ $=10\cdot 15\cdot 56=8400$

Total number of ways: $4200+11200+8400=23800$

Given this equals $100k$: $100k=23800$ $k=238$

Comparison with stored answer:

Stored correct answer = $238$.

Our derived answer matches it.