Combinatorics question, combinations involved in filling up of vacancies. To fill $12$ vacancies, there are $25$ candidates of which $5$ are from
  scheduled castes. If $3$ vacancies are reserved for scheduled caste
  candidates while the rest are open to all, find the number of ways in
  which selections can be made?


According to me it should be $^5C_3×^{22}C_9$ because we first find the combinations for scheduled castes then multiply it with the combinations for all the remaining candidates (including the scheduled castes).

However, answer given is $10 \times ^{20}C_9+ 5 \times ^{20}C_8 + ^{20}C_ 7$.

Why is my reasoning incorrect? 

Edit: I have realised my error, now how do I eliminate it and get the right answer?
The reasoning is incorrect because you're counting multiple times. Let's focus only on the "scheduled caste" section at the moment. Call the "scheduled caste" members A, B, C, D, E. These two cases are the same:


A, B, C make it through reservation and D, E make it through an open slot; rest same
A, D, E make it through reservation and B, C make it through an open slot; rest same


You have considered them as distinct. 

This would be clearer with a simpler problem: if you would get rid of all non-"scheduled" members make a selection of five from the "scheduled" group, with three reserved and two open. There is in effect only one way to do so, but with your method it is 5C3 * 2C2.

EDIT: To eliminate OP's error, just notice the simple fact that the number of people making it is $3$, $4$ or $5$ from the "scheduled" community. If we separated the entry process into the two classes reserved and non-reserved, we'd be counting multiple times because it doesn't matter what process a "scheduled" person is selected by as long as the selections are the same. Instead, we separate people into classes scheduled and non-scheduled. There are three cases: 3 from "scheduled", rest from open only; 4 from "scheduled", rest from open only and 5 from "scheduled", rest from open only.

So  we simply calculate $\binom{5}3 \cdot \binom{20}{9} + \binom{5}4 \cdot \binom{20}8 + \binom{5}5 \cdot \binom{20}7$ and we're done.
There are more candidates than vacancies.  The way the answer is structured is that possible ways the selection could be made is to fill the 3 vacancies from 5 SC, and the remaining 9 from Non-SC, 4 SC are placed and the remaining 8 from Non_SC and all 5 Scs are placed and the remaining 7 from Non-SC.  I would kind of agree with you that the 3 vacancies are filled from 5 SCs and the remaining nine vacancies are filled with 22 open candidates.  In the textbook solution, it assumes that all 5 SC candidates are placed which is not mentioned in the problem.