Chapter 11, Problem 13E

Solution

To find: The number of ways in which 3 items could be chosen for the first dinner, second dinner and fourth dinner.

The number of ways are220, 84, 20 and 1 respectively.

Given information:

Number of food items=12

Concept used:

The formula to compute the combination rule is:

  $\begin{array}{l}C\left(n,r\right)=\frac{n!}{r!\left(n-r\right)!}\\ \text{Where,}\\ n\text{ is the number of items}\\ r\text{ is the number of items to be chosen}\end{array}$

Calculation:

The number of ways in which 3 items could be chosen for the first dinner could be calculated as:

  $\begin{array}{c}C\left(12,3\right)=\frac{12!}{3!\left(12-3\right)!}\\ =\frac{12\times 11\times 10\times 9!}{9!\times 3\times 2\times 1}\\ =220\end{array}$

Number of items left for second dinner are:

  $\begin{array}{c}\text{Number of items left=}12-3\\ =9\end{array}$

The number of ways in which 3 items could be chosen for the second dinner could be calculated as:

  $\begin{array}{c}C\left(9,3\right)=\frac{9!}{3!\left(9-3\right)!}\\ =\frac{9\times 8\times 7\times 6!}{6!\times 3\times 2\times 1}\\ =84\end{array}$

Number of items left for the third dinner are:

  $\begin{array}{c}\text{Number of items left=9}-3\\ =6\end{array}$

The number of ways in which 3 items could be chosen for the third dinner could be calculated as:

  $\begin{array}{c}C\left(6,3\right)=\frac{6!}{3!\left(6-3\right)!}\\ =\frac{6\times 5\times 4\times 3!}{3!\times 3\times 2\times 1}\\ =20\end{array}$

Number of items left for the fourth dinner are:

  $\begin{array}{c}\text{Number of items left=}6-3\\ =3\end{array}$

The number of ways in which 3 items could be chosen for the fourth dinner could be calculated as:

  $\begin{array}{c}C\left(3,3\right)=\frac{3!}{3!\left(9-3\right)!}\\ =\frac{3!}{3!\times 0!}\\ =1\end{array}$

The required number of ways are 220, 84, 20 and 1 respectively.