Chapter 11.7, Problem 10E

To determine

To calculate: The number of schedules the club can choose.

Answer to Problem 10E

The total number of schedules that can be chosen are $10$ .

Explanation of Solution

Given information:

The club wants to meet on $3$ of the evenings next week from Monday through Friday.

Calculation: To find the number of sets of $3$ evenings the club can choose from Monday through Friday, use combination. As a set of $3$ days are to be chosen from $\text{5}$ days(Monday to Friday), the number of sets of $3$ days can be calculated as $\text{5C3}$ .

To find the number of combinations of $n$ objects taken $r$ at a time, divide the number of permutations of $n$ objects taken $r$ at a time by $r!$

  $nCr=\frac{nPr}{r!}$   [Combination formula]

  $\Rightarrow 5C3=\frac{5P3}{3!}$   [Use combination formula]

  $=\frac{5\cdot 4\cdot 3}{3\cdot 2\cdot 1}$   [Write $5P3$ and $3!$ as a product]

  $=\frac{5\cdot \cancel{\stackrel{2}{4}}\cdot \cancel{3}}{\cancel{3}\cdot \cancel{\underset{1}{2}}\cdot 1}$   [Divide out common factors]

  $=10$   [Simplify]

Therefore, the number of combinations are $10$ .

To make a list of all possible combinations of $3$ evenings the club can choose from Monday through Friday use the letters $\text{A,B,C,D,E}$ to represent all the days from Monday to Friday. List all possible matchups. Then cross out any duplicates that represent the same matchup.

  $\begin{array}{l}\text{ABC ABD ABE ACD ACE ADE}\\ \cancel{\text{BAC}}\cancel{\text{BAD}}\cancel{\text{BAE}}\text{ BCD BCE BDE }\\ \cancel{\text{CAB}}\cancel{\text{CAD}}\cancel{\text{CAE}}\cancel{\text{CBD}}\cancel{\text{CBE}}\text{ CDE}\\ \cancel{\text{DAB}}\cancel{\text{DAC}}\cancel{\text{DAE}}\cancel{\text{DBC}}\cancel{\text{DBE}}\cancel{\text{DCE}}\\ \cancel{\text{EAB}}\cancel{\text{EAC}}\cancel{\text{EAD}}\cancel{\text{EBC}}\cancel{\text{EBD}}\cancel{\text{EDC}}\end{array}$

Therefore, there are $10$ possible combinations