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'Medal Ceremony' printed from http://nrich.maths.org/
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.View the previous week's solutionView the current weekly problem