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.

it could be a ..idk tho

angle addition postulate

2x=56

step-by-step explanation: