A camera shop stocks eight different types of batteries, one of which is type a7b. assume there are at least 30 batteries of eac

Question

2 years ago

3

A camera shop stocks eight different types of batteries, one of which is type a7b. assume there are at least 30 batteries of eac

h type.
a. how many ways can a total inventory of 30 batteries be distributed among the eight different types?
b. how many ways can a total inventory of 30 batteries be distributed among the eight different types if the inventory must include at least four a76 batteries?

Mathematics

2 answers:

Svetradugi [14.3K] · Answer 1 · 2023-11-07T09:37:36+03:00

Statistical Method

For A:

Given:

k = 30
n = 8

Solution:

where ! = factorial

the required number is:

C (30 + 8 - 1, 30) = C (37, 30)

=37! / (37 - 30)! (30)!
= 10295472

For B:

Given:

k = 26
n = 8

Solution:

the require number is:

C (26 + 8 -1, 26) = C (33,26)

= 33! / (33 - 26)! (26)!
= 4272048

The answers are 10295472 for (a) and 4272048 for (b)

taurus [48] · Answer 2 · 2023-11-07T10:43:30+03:00

How many ways can a total inventory of 30 batteries be distributed among the eight different types? 10,295,472

<h3>Further explanation</h3>

A camera shop stocks eight different types of batteries, one of which is type a7b. assume there are at least 30

Definition permutation (order is important)

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

Repetition allowed: $n^r$

Definition combination (order is not important):

No repetition: $C(n,r) = \left[\begin{array}{ccc}n\\r\end{array}\right] = \frac{n!}{r!(n-r)!}$

Repetition allowed: $C(n+r-1,r) = \left[\begin{array}{ccc}n+r-1\\r\\\end{array}\right] = \frac{(n+r-1)!}{r!(n-1)!}$

with $n! = n*(n-1)*...*2*1$

a. how many ways can a total inventory of 30 batteries be distributed among the eight different types?

The camera shop stocks 8 different types of batteries and there are at least 30 batteries of each kind.

a) We want to select r = 30 batteries from the $n=8$ kinds of batteries

$r=30\\n=8$

The order of the batteries doesn't matter, thus we should use a combination. Moreover, repetition is allowed as we can choose multiple pastries of the same kind.

$\left[\begin{array}{ccc}30+8-1\\30\\\end{array}\right] = \left[\begin{array}{ccc}37\\30\end{array}\right]$

$= \frac{37!}{30!(37-30)!} \\= \frac{37!}{30!7!} \\= \frac{37*36*35*34*33*32*31*30!}{30!*(7*6*5*4*3*2*1)}$

$n!=n*(n-1)*...*2*1$

$=\frac{37*36*35*34*33*32*31}{7*6*5*4*3*2*1} \\=10,295,472$

b. how many ways can a total inventory of 30 batteries be distributed among the eight different types if the inventory must include at least four a76 batteries?

Assuming that we first select 4 A76 batteries, we need to determine on how many ways we can select the remaining $r=30-4=26$ batteries from the $n=8$ kinds of batteries

The order of the batteries doesn't matter, thus we should use a combination. Moreover, the repetition is allowed as we can choose multiple pastries of the same kind.

$\left[\begin{array}{ccc}26+8-1\\26\\\end{array}\right] = \left[\begin{array}{ccc}33\\26\end{array}\right]$