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There are $6$ different students who could receive the first gold medal, then $5$ others for the second and $4$ remaining for the third. Therefore there are $6 \times 5 \times 4 = 120$ orders in which the medals can be presented. However, this counts each set of three people winning the medals in each of the six orders, so there are $120 \div 6 = 20$ sets of three people who could win gold.

For each of these, one of the remaining three people must win bronze and the others silver, so there are $20 \times 3 = 60$ ways in which the medals can be awarded.

This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.