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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 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?

Respuesta :

missbianca
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)


tifanihayyu

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

Further explanation

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: [tex]P(n,r) = \frac{n!}{(n-r)!}[/tex]

Repetition allowed: [tex]n^r[/tex]

Definition combination (order is not important):

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

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

with [tex]n! = n*(n-1)*...*2*1[/tex]

  • 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 [tex]n=8[/tex] kinds of batteries

[tex]r=30\\n=8[/tex]

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.

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

[tex]= \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)}[/tex]

[tex]n!=n*(n-1)*...*2*1[/tex]

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

  • 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 [tex]r=30-4=26[/tex] batteries from the [tex]n=8[/tex] 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.

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

[tex]= \frac{33!}{26!(33-26)!} \\= \frac{33!}{26!7!} \\= \frac{33*32*31*30*29*28*27*26!}{26!*(7*6*5*4*3*2*1)}[/tex]

[tex]n!=n*(n-1)*...*2*1[/tex]

[tex]=\frac{33*32*31*30*29*28*27*26!}{26!*(7*6*5*4*3*2*1} \\=4,272,048[/tex]

Learn more

  1. Learn more about inventory https://brainly.com/question/9223260
  2. Learn more about batteries https://brainly.com/question/8898675
  3. Learn more about a76 batteries https://brainly.com/question/11554915

Answer details

Grade:   5

Subject: math

Chapter:  camera shop stocks

Keywords:  stocks, camera, shop, batteries, inventory

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