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Counting analysis




Arragements


Let's consider the following set of digits Set= {1,2,3} Question: How many numbers of two digits can we obtain from this set?. The answer is: 12, 13, 21, 23, 31, and 32. In total, 6 numbers. The order of these numbers is NOT important. The arragement 13 and 31 , for example, are not the same. The question would be presented as: How many arrangements of two digits can we make from the three elements of the set Set? There are 6 manners to arrange two digits among three digits. In general, we process as follows, to arrange p objects among n objects: There is n possibilities to choose the first object. Once done, it remains (n-1) possibilities to choose the second object, Once done, it remains (n-2) possibilities to choose the third object, ... etc ... Once done, it remains (n-(r-1)) possibilities to choose the r th object. In a total, there is A = n x (n-1) x (n-2) x ... x (n-r+1) arrangements of r objects among n objects. The result A can be written: A = A x R/R where R = (n-r) x (n-r-1) x (n-r-2)... x 3 x x 2 x 1 A x R = n! R = (n-r)! Finally:
The number of arrangements of r objects among n objects is A(r, n) = n!/(n-r)!



  
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