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Topic: Foot Buddies (Read 1524 times) |
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ThudnBlunder
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Foot Buddies
« on: Jan 20th, 2009, 6:32pm » |
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Six colleges each send their two best runners to an athletics meeting. The 12 runners are randomly assigned to the six available lanes, with two runners in each lane. What is the probability that at least one college will have both of its runners assigned to the same lane?
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THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
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balakrishnan
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Re: Foot Buddies
« Reply #1 on: Jan 21st, 2009, 2:14am » |
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The number of ways in which 2N runners(2 from N teams each) shall be distributed in the N lanes such that atleast one college has both its runners assigned to the same lane is given by(using simple inclusion-exclusion)
sum(i=1,N,(-1)^(i-1)*(binomial(N,i))^2*i!*2^i*(2*N-2*i)!)
and hence the probability is
sum(i=1,N,(-1)^(i-1)*(binomial(N,i))^2*i!*2^i*(2*N-2*i)!)/(2*N)!
For N=6, the probability happens to be 871/2079
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