1. Assuming we're not concerned with the order of finish, the number is the number of ways 5 can be chosen from 43.
... C(43, 5) = 43!/(5!·(43-5)!) = (43·42·41·40·39)/(5·4·3·2·1) = 962,598
2. There are 102 choose 2 = C(102, 2) = 102*101/(2*1) = 5,151 ways to pair one student with another.
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There are more than 2.75×10⁸⁰ ways the class of 102 can be divided into pairs of students. That number is 102!/(51!·2⁵¹).
