Acatering service offers 9 appetizers 8 main courses and 3 desserts. a customer is to select 6 appetizers 5 main courses and 2 desserts for a banquet. in how many ways can this be done?

Explanation:

We'll be using the n C r combination function. To make the notation a bit easier to deal with, I will write "C(n,r)" instead of "n C r".

The formula is

C(n,r) = (n!)/(r!*(n-r)!)

where the exclamation marks represent factorials.

A factorial is where you start with a positive integer, and count down to 1 multiplying all along the way.

Examples:

5! =           5*4*3*2*1

8! = 8*7*6*5*4*3*2*1 = 120

Note how the string "5*4*3*2*1" is in both 5! and 8!

We can say 8! = 8*7*6*5!

Because we can replace the "5!" at the end with "5*4*3*2*1" later if we wanted. This strategy is used to help find a shortcut to simplification.

We have n = 9 appetizers and r = 6 items we can select from this pool.

C(n,r) = (n!)/(r!*(n-r)!)

C(9,6) = (9!)/(6!*(9-6)!)

C(9,6) = (9!)/(6!*3!)

C(9,6) = (9*8*7*6!)/(6!*3*2*1)

C(9,6) = (9*8*7)/(3*2*1) the "6!" terms canceled out

C(9,6) = 504/6

C(9,6) = 84

There are 84 ways to choose six appetizers from the pool of nine available

Repeat those steps for the main courses. Use n = 8 and r = 5 this time.

C(n,r) = (n!)/(r!(n-r)!)

C(8,5) = (8!)/(5!*(8-5)!)

C(8,5) = (8!)/(5!*3!)

C(8,5) = (8*7*6*5!)/(5!*3*2*1)

C(8,5) = (8*7*6)/(3*2*1)

C(8,5) = (336)/(6)

C(8,5) = 56

There are 56 ways to choose five main course meals from the pool of eight available

Then do the same for the desserts. Use n = 3 and r = 2.

C(n,r) = (n!)/(r!(n-r)!)

C(3,2) = (3!)/(2!*(3-2)!)

C(3,2) = (3!)/(2!*1!)

C(3,2) = (3*2*1)/(2*1*1)

C(3,2) = 6/3

C(3,2) = 3

There are 3 ways to choose two desserts from the pool of three available

The last step is to multiply all these results:

84*56*3 = 14112

This is the number of ways to select all of the items given the restrictions listed. The order does not matter.

a.) list the factors of 40.

1/2/4/5/8/10/20/40

b.) list the factors of 40 that represent the number of cans that could be on each shelf.

5

c.) how many different arrangements of cans are possible?

just one because if you put more on one shelf you will go way over 40 and the cabinet will probably break.

d.) what are the possible arrangements of the cans?

put 5 cans on each self to reach 40 cans in total.

e.) which arrangement uses all the shelves?

5 because you can put 5 cans on each shelf, and since there are 8 shelfs you are equally going to get 40 cans.