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# How many different ways can 30 runners place in an Olympic

qualifying marathon?...
How many different ways can 30 runners place in an Olympic qualifying marathon?
Answered in 2 minutes by:
2/10/2012
Sandhya_sharma, Master's Degree
Category: Math Homework
Satisfied Customers: 3,715
Experience: I hold M.Sc and M.Phil degrees in math and have several years of teaching experience.
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Hi,

Welcome to Justanswer!

How many different ways can 30 runners place in an Olympic qualifying marathon?

Solution:

There are 30 choices for first runner, then 29 for second and then so on

Number of ways = 30*29*28....*1(upto 1) = 30!

Answer: 30!

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Customer reply replied 5 years ago
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?
Chirag, Master's Degree
Category: Math Homework
Satisfied Customers: 12,192
Experience: Excellent Tutor with a long teaching experience at different levels.
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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.
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dhouse1940, Master's Degree
Category: Math Homework
Satisfied Customers: 676
Experience: BS mathematics, MS biostatistics, 35+ yrs designing & analyzing biological experiments.
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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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Customer reply replied 5 years ago

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?

Is there a way to tip more for the answer since you are helping me not only with the answer but understanding it?

Yes to first question.
Yes to second question (I checked math)

Thanks, XXXXX XXXXX leave a bonus and it would be appreciated.
dhouse1940, Master's Degree
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Experience: BS mathematics, MS biostatistics, 35+ yrs designing & analyzing biological experiments.
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