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Show that \(\left( {\begin{array}{*{20}{l}}{\rm{n}}\\{\rm{k}}\end{array}} \right){\rm{ = }}\left( {\begin{array}{*{20}{c}}{\rm{n}}\\{{\rm{n - k}}}\end{array}} \right)\). Give an interpretation involving subsets.

Short Answer

Expert verified

There are same number of subsets of those sizes.

Step by step solution

01

Definition

The term "probability" simply refers to the likelihood of something occurring. We may talk about the probabilities of particular outcomes—how likely they are—when we're unclear about the result of an event. Statistics is the study of occurrences guided by probability.

02

Explanation

We have

\(\left( {\begin{array}{*{20}{l}}{\rm{n}}\\{\rm{k}}\end{array}} \right){\rm{ = }}\frac{{{\rm{n!}}}}{{{\rm{k!(n - k)!}}}}\)

and also

\(\begin{array}{c}\left( {\begin{array}{*{20}{c}}{\rm{n}}\\{{\rm{n - k}}}\end{array}} \right){\rm{ = }}\frac{{{\rm{n!}}}}{{{\rm{(n - k)!(n - (n - k))!}}}}\\{\rm{ = }}\frac{{{\rm{n!}}}}{{{\rm{(n - k)!k!}}}}\end{array}\)

from which we have the equality

\(\left( {\begin{array}{*{20}{l}}{\rm{n}}\\{\rm{k}}\end{array}} \right){\rm{ = }}\left( {\begin{array}{*{20}{c}}{\rm{n}}\\{{\rm{n - k}}}\end{array}} \right)\)

The number of unordered subsets size \({\rm{k}}\) is the same as the number of unordered subsets size \({\rm{(n - k)}}\). This is true because for every subset of \({\rm{k}}\) elements, the other (left) \({\rm{(n - k)}}\) elements create a subset of size \({\rm{(n - k)}}\), this is why there are same number of subsets of those sizes.

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