Predicting final standings in combinatorics 
There are 10 athletes participating in a competition. How many possible final
  standings, in respect to the first three places, would one have to
  predtict to be sure that at least one of them:  a) without taking
  order into account b) taking order into account is correct?

So the answers for these in my notes are $3!+3\cdot2\cdot8+3\cdot9\cdot8$ for a) and ${3\choose1}{7\choose2}+{3\choose2}{7\choose1}+{3\choose3}$ for b).
Why is that so? I mean - if we have to make a correct prediction without thinking about the order, don't we have to simply calculate ${10\choose3}$? And as for the prediction with order, why isn't the answer the same as the previous but multiplied by all the permutations of 3-element set so ${10\choose3}\cdot3!$ ?
EDIT: OK, it seems the interpretation of the problem seems different cause the answer for b) in the notes gives the predictions we'd have to make to correctly predict exactly one, exactly two or exactly three of the people on the podium. The answer for a seems more mysterious to me so if someone could help me get it, it'd be great :) 
 A: The question is poorly phrased and difficult to interpret.  I've given it my best shot, but do not get the same answers.

Basically: You select 3 of 10 items into the predictions ordered lists, the race then independently selects 3 of the same 10 items into the standings ordered list.
(A) Find: How many permutations of predictions can there be that have at least one common element to any given standings, when the placement order does not have to match.

Count the ways to: choose all 3 in, choose 2 in and 1 out, choose 1 in and 2 out.  Shuffle them.
$$3!\times\left({3\choose 3}+{3\choose 2}{7\choose 1}+{3\choose 1}{7\choose 2}\right) \\ = \left(3!+3\times 2\times 3\times 7+9\times 7 \times 6\right)$$

(B) Find: How many permutations of predictions can there be that have at least one common element to any given standings, when the placement order must match.

Count the ways to: Choose all three placed correctly. Choose two correct places and one incorrect athlete.  Choose one correct place, and two incorrect athletes (in order).
$${3\choose 3}+{3\choose 2}\times 7+{3\choose 1}\left(9\times 8-2\times 8+1\right)$$
(Note on the last term: using inclusion/exclusion: there are 9 items to choose into two places, but 2 of them have forbidden placement. $9\times 8$ permutations in total, $2\times 8$ where at least one item is in its forbidden place, and $1$ where both are.)
