Discrete math problem college level question

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The problem involves determining the number of non-crossing handshake arrangements among 2n people seated at a round table, denoted as S_n. To find S_10, the discussion suggests exploring smaller values of n to identify a pattern. The formula derived for the total number of arrangements is S_n = n(n-1)/2, accounting for the fact that handshakes in opposite directions are considered identical. By substituting n=10 into the formula, it is calculated that S_10 equals 45. Thus, there are 45 distinct ways for 10 people to shake hands without crossing.
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Suppose 2n people sit on a round table and are shaking hands in
pairs. Suppose that etiquette is observed and no 2 shakes cross. Let
S_n be the number of possible shaking hands arrangements of this sort.

Determine S_10.
 
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I recommend trying to work this out for n=2, 3, and 4 and then find a pattern.

You'll need to find a formula for the sum of the first k whole numbers which is probably in your book somewhere.
 


To solve this problem, we can use the concept of permutations and combinations. Since each person can only shake hands with one other person at a time, the total number of handshakes that can occur is n (the number of people) multiplied by (n-1) (the number of potential handshakes for each person). This can be written as n(n-1).

However, this calculation includes both clockwise and counterclockwise handshakes, which are considered the same in this scenario. Therefore, we need to divide the total number of handshakes by 2 to account for this repetition. This gives us the formula:

S_n = n(n-1)/2

Substituting n=10 into this formula, we get:

S_10 = 10(10-1)/2 = 45

Therefore, there are 45 possible ways for 10 people to shake hands in pairs around a round table while observing etiquette and without any crossed handshakes.
 
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