Chapter 6.1, Problem 50PS

Solution

To find: The number of different combinations of arcade games.

60460 different combinations.

Given:

An arcade has 20 different arcade games and you want to play at least 14 of them.

Calculation:

The number of ways of playing 14 games from 20 different games is,

  $\begin{array}{c}C{}_{14}=\frac{20!}{\left(20-14\right)!14!}\\ =\frac{20\cdot 19\cdot 18\cdot 17\cdot 16\cdot 15\cdot \cancel{14!}}{6!\cancel{14!}}\\ =\frac{27907200}{720}\\ =38760\end{array}$

The number of ways of playing 15 games from 20 different games is,

  $\begin{array}{c}C{}_{15}=\frac{20!}{\left(20-15\right)!15!}\\ =\frac{20\cdot 19\cdot 18\cdot 17\cdot 16\cdot \cancel{15!}}{5!\cancel{15!}}\\ =\frac{1860480}{120}\\ =15504\end{array}$

The number of ways of playing 16 games from 20 different games is,

  $\begin{array}{c}C{}_{16}=\frac{20!}{\left(20-16\right)!16!}\\ =\frac{20\cdot 19\cdot 18\cdot 17\cdot \cancel{16!}}{4!\cancel{16!}}\\ =\frac{116280}{24}\\ =4845\end{array}$

The number of ways of playing 17 games from 20 different games is,

  $\begin{array}{c}C{}_{17}=\frac{20!}{\left(20-17\right)!17!}\\ =\frac{20\cdot 19\cdot 18\cdot \cancel{17!}}{3!\cancel{17!}}\\ =\frac{6840}{6}\\ =1140\end{array}$

The number of ways of playing 18 games from 20 different games is,

  $\begin{array}{c}C{}_{18}=\frac{20!}{\left(20-18\right)!18!}\\ =\frac{20\cdot 19\cdot \cancel{18!}}{2!\cancel{18!}}\\ =\frac{380}{2}\\ =190\end{array}$

The number of ways of playing 19 games from 20 different games is,

  $\begin{array}{c}C{}_{19}=\frac{20!}{\left(20-19\right)!19!}\\ =\frac{20\cdot \cancel{19!}}{1!\cancel{19!}}\\ =20\end{array}$

The number of ways of playing 20 games from 20 different games is,

  $\begin{array}{c}C{}_{20}=\frac{20!}{\left(20-20\right)!20!}\\ =\frac{\cancel{20!}}{0!\cancel{20!}}\\ =1\end{array}$

Hence, the number of ways of playing at least 14 games from 20 different games is,

  $38760+15504+4845+1140+190+20+1=60460$ .