Question

resent a combinatorial argument for this identity by considering a set of n people and determining, in two ways, the number of possible selections of a committee of any size and a chairperson for the committee.

Answer 1

The committee can be selected by combinatorial argument  in

$\sum ^{n}_{k=1} k(^n_k).$ ways.

A counting-based argument is known as a combinatorial argument or combinatorial proof. This line of reasoning has previously been used, for instance in the section on Stirling numbers of the second sort.

By initially selecting k individuals from our group of n, we can then choose one of those k individuals to serve as the committee's chairperson.

A number of methods for completing the first task, k methods for completing the second task, and so on. ways to create a k-member committee with a chairperson.

$\sum ^{n}_{k=1} k(^n_k).$ is the number of methods to construct a committee with a chairman of size less than or equal to n can be found by adding up over 1≤k≤n.

A committee of size less than or equal to n can also be formed with a chairperson by selecting the chairperson first, followed by the members of the committee. The chairperson can be chosen from among n options. The picker has two options for the remaining n-1 individuals: to include them or not. We therefore have n options for the chairperson, 2 options for the following, 2 options for the following, etc. These can be multiplied together to give us $n2^{n-1}$, which is a proof of the identity.

To learn more about combinatorial argument:

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