But how would you right that out in a numerical answer.... also how would you determine the different ways three medalists (gold, silver, and bronze) can be chosen from the whole field of 30 runners?

Hi, can you please post the complete question? Will the fastest 8 runners qualify for the olympics? And you want to find the number of ways the 3 winners can be selected? Please clarify.

30! is correct and the actual number is XXXXX given because it is huge and tedious to compute. The answer is 30! = 30*29*28*27*26*25*24*23*22*21*20*19*18*17*16*15*14*13*12*11*10*9*8*7*6*4*3*2*1

The different combinations of 3 medalists from 30 runners is

nCx = n! / [x!*(n – x)!] where a! = a*(a–1)*(a–2)*…*1. So, 30! / [3!*27!] = 30*29*28*27! /(3!*27!) = 30*29*28/ 3! = 5*29*28 = 4060 For each combination there are 3*2*1 = 6 possible orderings So the total number of different gold, silver, bronze orderings is 4060*6 = 24,360 Or, simply 30*29*28 = 24,360

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Just to make sure I understand the difference between permutation and combination. You would use the equation for a permutation to establish gold, silver, and bronze because order does matter. However, if you were going to establish the top eight fastest runners, you would use the cobination formula because the top eight will always be the top eight and order would not matter. Correct?

So if I were going to determine the number of different ways the eight fastest runners could be chosen from the field of 30, my formula to work out would be 30!/8!22! - which breaks down to 5,852,925 ways, correct?

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