Chapter 6, Problem 5REAE

Solution

To Find: The different combinations of ice cream flavors that can be sampled.

The different combinations are $32,192$

Given information: An ice cream vendor sells $15$ flavors of ice cream. $4$ flavors need to be sampled at least.

Calculation:

The combinations can be written as:

  $C{}_4+C{}_5+C{}_6+C{}_7+C{}_8+C{}_9+C{}_{10}+C{}_{11}+C{}_{12}+C{}_{13}+C{}_{14}+C{}_{15}$

This combination can be solved as:

  $C{}_r=\frac{n!}{r!(n-r)!}$

Where:

n is the number of ice cream.

r is the unique arrangement of ice cream.

Hence, this given combinations can be solved as:

  $\begin{array}{l}C{}_4=\frac{15!}{4!(15-4)!}\\ C{}_4=\frac{15\times 14\times 13\times 12\times 11!}{4\times 3\times 2\times 1\times 11!}\\ C{}_4=1365\end{array}$

  $\begin{array}{l}C{}_5=\frac{15!}{5!(15-5)!}\\ C{}_5=\frac{15\times 14\times 13\times 12\times 11\times 10!}{5\times 4\times 3\times 2\times 1\times 10!}\\ C{}_5=3003\end{array}$

  $\begin{array}{l}C{}_6=\frac{15!}{6!(15-6)!}\\ C{}_6=\frac{15\times 14\times 13\times 12\times 11\times 10\times 9!}{6\times 5\times 4\times 3\times 2\times 1\times 9!}\\ C{}_6=5005\end{array}$

  $\begin{array}{l}C{}_7=\frac{15!}{7!(15-7)!}\\ C{}_7=\frac{15\times 14\times 13\times 12\times 11\times 10\times 9\times 8!}{7\times 6\times 5\times 4\times 3\times 2\times 1\times 8!}\\ C{}_7=6435\end{array}$

  $\begin{array}{l}C{}_8=\frac{15!}{8!(15-8)!}\\ C{}_8=\frac{15\times 14\times 13\times 12\times 11\times 10\times 9\times 8!}{8!\times 7\times 6\times 5\times 4\times 3\times 2\times 1}\\ C{}_8=6435\end{array}$

  $\begin{array}{l}C{}_9=\frac{15!}{9!(15-9)!}\\ C{}_9=\frac{15\times 14\times 13\times 12\times 11\times 10\times 9!}{9!\times 6\times 5\times 4\times 3\times 2\times 1}\\ C{}_9=5005\end{array}$

  $\begin{array}{l}C{}_{10}=\frac{15!}{10!(15-10)!}\\ C{}_{10}=\frac{15\times 14\times 13\times 12\times 11\times 10!}{10!\times 5\times 4\times 3\times 2\times 1}\\ C{}_{10}=3003\end{array}$

  $\begin{array}{l}C{}_{11}=\frac{15!}{11!(15-11)!}\\ C{}_{11}=\frac{15\times 14\times 13\times 12\times 11!}{11!\times 4\times 3\times 2\times 1}\\ C{}_{11}=1365\end{array}$

  $\begin{array}{l}C{}_{12}=\frac{15!}{12!(15-12)!}\\ C{}_{12}=\frac{15\times 14\times 13\times 12!}{12!\times 3\times 2\times 1}\\ C{}_{12}=455\end{array}$

  $\begin{array}{l}C{}_{13}=\frac{15!}{13!(15-13)!}\\ C{}_{13}=\frac{15\times 14\times 13!}{13!\times 2\times 1}\\ C{}_{13}=105\end{array}$

  $\begin{array}{l}C{}_{14}=\frac{15!}{14!(15-14)!}\\ C{}_{14}=\frac{15\times 14!}{14!\times 1}\\ C{}_{14}=15\end{array}$

  $C{}_{15}=1$

Now, Substitute the value:

  $\begin{array}{l}C{}_4+C{}_5+C{}_6+C{}_7+C{}_8+C{}_9+C{}_{10}+C{}_{11}+C{}_{12}+C{}_{13}+C{}_{14}+C{}_{15}\\ 1365+3003+5005+6435+6435+5005+3003+1365+455+105+15+1\\ 32,192\end{array}$