Posted by: anthonygenello | November 20, 2008

Math 152 helps you get jobs…

…maybe. Last week, I had an interview with a quantitative trading firm. I was asked to solve a probability problem that’s directly related to this course. The question was:

Five friends are seated at a dinner table randomly. If no one shares a birthday, what is the probability that they will be seated in order of age (either decreasing or increasing) around the table?

We came across a similar problem in the beginning of the course when discussing cycle structures. You’ll probably remember that the number of distinct cycle structures for a set of 5 elements is 4! = 24. Now how many of these correspond to decreasing order of ages? 1. Increasing order of ages? 1. So 2 out of 24 arrangements work. Therefore the probablility we are looking for is 1/12. In general, for n friends, the probability of correct age ordering is

\frac{2}{(n-1)!}

That’s one way to solve the problem, but we can also apply another math 152 concept to solve this problem – permutations. Assume without loss of generality that the 5 friends sit down in order of age and then perform a random permutation by switching seats. How many permutations are possible. This is simply

|S_5| = 5! = 120

How many permutations will lead to another correct age ordering of the friends? We know that any permutation group of a polygon will preserve a cycle structure (if we are indifferent between forward and backward ordering). Thus the permutations of a pentagon’s symmetry group will give us correct orderings. Remember that for a pentagon, we have one identity, four rotations, and five reflections. Thus, the following number of permutations will work:

|D_5|=1+4+5=10

Since 10 out of 120 permutations will lead to the correct age ordering, the probability of 5 friends sitting in order of ages is 10/120 = 1/12. We get the same result either way!

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  1. […] Math 152 helps you get jobs… […]


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