Definition permulatin (oder is important)

No repetition allowed: \(\displaystyle{P}{\left({n},{r}\right)}={\frac{{{n}!}}{{{\left({n}-{r}\right)}!}}}\)

Repetition allowed: \(\displaystyle{n}^{{r}}\)

Definition combination(order is not important):

No repetition allowed: \[(\begin{matrix}n\\ r \end{matrix})=\frac{n!}{r!(n-r)!}\]

Repetition allowed: \[C(n+r-1, r)=( \begin{matrix}n+r-1 \\ r \\ \end{matrix} ) =\frac{(n+1-1)!}{r!(n-1)!}\]

with \(\displaystyle{n}\ne{n}\cdot{\left({n}-{1}\right)}\cdot\cdots{c}\cdot{2}\cdot{1}\)

Solution

The camera shop stokcs8 different types of batteries and there are at least 30 batteries of each kind

(a) We want to select r=30 batteries frrom the n=8 kinds of batteries,

\(\displaystyle{r}={30}\)

\(\displaystyle{n}={8}\)

The order of the batteries doesn't matter, thus we shuold use a combination . Moreover, repetition is allowed as

we can choose mulptiple pastries of the same kind.

\[( \ \begin{matrix} 30+8-1 \\ 30 \\ \end{matrix}\ )=( \ \begin{matrix}37 \\30 \\ \end{matrix}\ )\]

\(\displaystyle={\frac{{{37}!}}{{{30}!{\left({37}-{30}\right)}}}}\)

\(\displaystyle={\frac{{{37}!}}{{{30}!{7}!}}}\)

\(\displaystyle={\frac{{{37}\cdot{36}\cdot{35}\cdot{34}\cdot{33}\cdot{32}\cdot{31}\cdot{30}!}}{{{30}!\cdot{\left({7}\cdot{6}\cdot{5}\cdot{4}\cdot{3}\cdot{2}\cdot{1}\right)}}}}\)

\(\displaystyle={\frac{{{37}\cdot{36}\cdot{35}\cdot{34}\cdot{33}\cdot{32}\cdot{31}}}{{{7}\cdot{6}\cdot{5}\cdot{4}\cdot{3}\cdot{2}\cdot{1}}}}\)

\(\displaystyle={10},{295},{472}\)

(b) Assuming that we first select 4 A76 batteries, we need to determine on how many ways we can select the

remaining \(\displaystyle{r}={30}-{4}={26}\) batteries from the \(\displaystyle{n}={8}\) kinds of batteries.

The order of the batteries doesn't matter, thus we should use a combination. Moreover, repetition is allowed as

we can choose muitiple pasrties of the same kind.

\[(\ \begin{matrix} 26+8-1 \\ 26 \\ \end{matrix}\ ) =(\ \begin{matrix} 33 \\ 26 \\ \end{matrix}\ )\]

\[=(\ \begin{matrix} 33! \\ 26!(33-26) \\ \end{matrix}\ )\]

\[=(\ \begin{matrix} 33! \\ 26!7! \\ \end{matrix}\ )\]

\(\displaystyle={\frac{{{33}\cdot{32}\cdot{31}\cdot{30}\cdot{29}\cdot{28}\cdot{27}\cdot{26}!}}{{{26}!\cdot{\left({7}\cdot{6}\cdot{5}\cdot{4}\cdot{3}\cdot{2}\cdot{1}\right)}}}}{n}\ne{n}\cdot{\left({n}-{1}\right)}\cdot\cdots\cdot{2}\cdot{1}\)

\(\displaystyle={\frac{{{33}\cdot{32}\cdot{31}\cdot{30}\cdot{29}\cdot{28}\cdot{27}!}}{{{7}\cdot{6}\cdot{5}\cdot{4}\cdot{3}\cdot{2}\cdot{1}}}}\)

\(\displaystyle={4},{272},{048}\)

(c) By part (a) and (b) we know that 4,272,048 of the 10,295,472 selections of the batteries result in at least 4

A76 batteries and thus 10,295,472-4,272,048=6,023,424 selections of thirty batteries result in at most 3

A76 batteries.