Respuesta :
Answer: 14112
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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.
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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
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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
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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
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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.
