# A camera shop stocks eight different types of batteries, one of which

David Lewis 2021-12-17 Answered
A camera shop stocks eight different types of batteries, one of which is type A7b.Assume there are at least 30 batteries.
of each type.a. How many ways can a total inventory of 30 batteries be distributed among the eight different types? b.
How many way can a total inventory of 30 batteries be distributed among the eight different types of the inventory must
include at least four A76 batteries?c. How many ways can a total inventory of 30 batteries be distributed among the eight
different types of the inventory includes at most three A7b batteries

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Juan Spiller
Denitions
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.
###### Not exactly what you’re looking for?
stomachdm
a) No of ways can a total inventory of
30 batteries be distributed among the
eight different types
$=\ ( \begin{matrix} 30+8-1 \\ 30\\ \end{matrix}\ )=\ ( \begin{matrix} 37 \\ 30\\ \end{matrix}\ )$
$$\displaystyle={\frac{{{37}!}}{{{30}!{7}!}}}={\frac{{{37}\times{36}\times{35}\times{34}\times{33}\times{32}\times{31}}}{{{7}\times{6}\times{5}\times{4}\times{3}\times{2}\times{1}}}}$$
$$\displaystyle{10295472}$$
b) The number of ways can a total inventory
of 30 batteries be distributed among 8
different types if the inventory must include
at least 4 A76 batteries
$=\sum_{i=1}^3 (\ \begin{matrix} (30-k)+7-1 \\ 30-k \\ \end{matrix}\ )=(\ \begin{matrix} 36 \\ 30 \\ \end{matrix}\ )+(\ \begin{matrix} 35 \\ 29 \\ \end{matrix}\ )(\ \begin{matrix} 34 \\ 28 \\ \end{matrix}\ )+(\ \begin{matrix} 33 \\ 27 \\ \end{matrix}\ )$
$$\displaystyle={1947792}+{1623160}+{1344904}+{1107568}$$
$$\displaystyle={6},{023},{424}$$
nick1337

(a)
To calculate:
The number of ways a batch of 30 batteries can be distributed among eight
categories.
Formula used:
The namber of ways to select r number of elements from a set of n elements is
given by:
$(\ \begin{matrix} r+n-1\\ r\\ \end{matrix}\ )=\frac{(r+n-1)!}{(r+n-1-r)!\times r}$
Calculation
It is nedded to select an inventory of 30 individuals from eight diffeent types and
it is possible to select all units from one types or miss one or few types in the
selection.
Let n=8, r=30.
$[\ \begin{matrix} 30+8-1\\ 30\\ \end{matrix}\ ]=\frac{(30+8-2)!}{(30+8-1-30)!\times 30!}$
$$=10,295,472$$
(b)
To find :
Thenumber of possibilities to select inventore of 30 batteries  representing all eught
categories including at least 4 batteries form category A76.
Calculation
In part(a) we have obtained the number is shows that  possible ways to select 30
batteries. By subtracting the number of ways to select 30 batteries includung ar
most 3 units of A76 from this total count, the number of ways that can be select a
batch of 30 batteries with at least 4 batteries of A76 can be found.
Now substitute n=7, r=29.
[$\ \begin{matrix} 29+7-1\\ 29\\ \end{matrix}\ ]=\frac{(29+7-1)!}{(29+7-1-29)!\times 29!}$
$$=1M623,160$$
Then, the selection of 2 units from A776 is substituting n=7, r=28.
$[\ \begin{matrix} 28+7-1\\ 28\\ \end{matrix}\ ]=\frac{(28+7-1)!}{(28+7-1-28)!\times 28!}$
$$=1,344,904$$
Then, the selection of 3 units from A76 is substitutiong n=7, r=27.
$[\ \begin{matrix} 27+7-1\\ 27\\ \end{matrix}\ ]=\frac{(27+7-1)!}{(27+7-1-28)!\times 27!}$
If there is no A76 batteries then n=7, r=30.
$[ \ \begin{matrix} 30+7-1 \\ 30 \\ \end{matrix}\ ]=\frac{(30+7-1)!}{(30+7-1-30)!\times30!}$
$$=1,947,792$$
Thus, the total number of possible ways to selecting 30 batteries,
$$10,205,472-(1,623,160+1,344,904+1,107,568+1,947,792)$$
$$=4,272,048$$
(c)
To find:
The number of possible ways to select a batch of 30 batteries including at most 3
batteries from the type AT6
Calculation:
The number of possible ways to select a batch of 30 batteries includung at most 3
batteries from the type A76 can be calculated  by subtract nymber of possible ways
to selecting 30 batteries including at least 4 of A76 from total possibility.
That is
$$10,295,472-4,272,048=6,023,242$$

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