Question

A camera shop stocks eight different types of batteries, one of which is type A76. Assume there are at least 30 batteries of each type.

Required:
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?
c. How many ways can a total inventory of 30 batteries be distributed among the eight different types if the inventory includes at most three A7b batteries?

Answer 1

a. 10295472 ways

b. 4272048 ways

c. 6023424 ways

Explanation:

Given that:

Camera shop stocks ----- 8 different types of batteries

one of which is ---- A76

Assume that 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.

The number of ways a total inventory of 30 batteries be distributed is :

`= \left \{ {{30+8-1} \atop {30}} \right. \}`

`= \left \{ {{37} \atop {30}} \right. \}`

`=(37!)/(30! *7!)`

`= (37*36*35*34*33*32*31*30!)/(30!*7*6*5*4*3*2*1)`

`= (37*36*35*34*33*32*31)/(7*6*5*4*3*2*1)`

= 10295472 ways

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?

If we must include 4 A76 batteries; then the number of ways a total inventory of 30 batteries can be distributed among eight different types of batteries will be:

30 - 4 = 26 batteries

Now;

`= \left \{ {{26+8-1} \atop {26}} \right. \}`

`= \left \{ {{33} \atop {26}} \right. \}`

`=(33!)/(26! \ \ 7!)`

`=(33*32*31*30*29*28*27*26!)/(26! \ * \ 7*6*5*4*3*2*1)`

`=(33*32*31*30*29*28*27)/( \ 7*6*5*4*3*2*1)`

= 4272048 ways

c. If we must include at most three A7b batteries. the number of ways that a total inventory of 30 batteries can be distributed among eight different types of inventory is:

`= \sum \limits ^3 _(x=0) \left \{ {{(30-x)+7-1} \atop {30-x}} \right. \} \\ \\ \\ = \sum \limits ^3 _(x=0) \left \{ {{(30-0)+7-1} \atop {30-0}} \right. \} = (^(36)_(30))+(^(35)_(29))+ (^(34)_(28))+ (^(33)_(27))`

`= (36!)/(30! * 6!) + (35!)/(29! * 6!) + (34!)/(28! * 6!) + (33!)/(27! * 6!)`

= 1947792 + 1623160 + 1344904 + 1107568

= 6023424 ways