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In how many ways can we select a chairperson, vice-chairperson and recorder from a group of 11 persons?

Four members from a 50-person committee are to be selected randomly to serve as chairperson, vice-chairperson, secretary, and treasurer. The first person selected is the chairperson; the second, the vice-chairperson; the third, the secretary; and the fourth, the treasurer. How many different leadership structures are possible?

In a group of 191 students, 10 are taking French, business, and music; 36 are taking French and business; 20 are taking French and music; 18 are taking business and music; 65 are taking French; 76 are taking business; and 63 are taking music. Use the Inclusion-Exclusion Principle for three finite sets to determine how many students are not taking any of the three courses.

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In how many ways can we select a chairperson, vice-chairperson, secretary, and treasurer from a group of 12 persons?

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We are interested in the number of 4 people from a set with 12 persons:

\begin{align*} n&=12 \\ r&=4 \end{align*}

The order in which we select the 4 people matters (as a different order of the people results in different chairperson, vice-chairperson, secretary and treasurer), thus we need to use the definition of a $\textbf{permutation}$ .

A $\textbf{r-permutation}$ of a set of elements is an ordering of the $r$ of the elements in the set in a row. The number of $r$ -permutations of a set of $n$ distinct objects is $P(n,r)=\frac{n!}{(n-r)!}$ .

\begin{align*} P(12,4)&=\frac{12!}{(11-4)!} \\ &=\frac{12!}{8!} \\ &=\frac{12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1} \\ &=12\cdot 11\cdot 10\cdot 9 \\ &=11,880 \end{align*}

Thus there are 11,880 ways to select a chairperson, vice-chairperson, secretary, and treasurer from a group of 12 persons.

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