Question 1161704
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Use the combination formula
*[Tex \Large _n C _r = \frac{n!}{r!*(n-r)!}]
Order does not matter because any dish does not outrank another.


We have 5 appetizers overall, and can only select 4 of them, so n = 5 and r = 4
*[Tex \Large _n C _r = \frac{n!}{r!*(n-r)!}]


*[Tex \Large _5 C _4 = \frac{5!}{4!*(5-4)!}]


*[Tex \Large _5 C _4 = \frac{5!}{4!*1!}]


*[Tex \Large _5 C _4 = \frac{5*4*3*2*1}{4*3*2*1*1}] Factorials are where we start with the given value, count down til we get to 1, multiplying all along the way. 


*[Tex \Large _5 C _4 = \frac{120}{24}]


*[Tex \Large _5 C _4 = 5]


There are 5 ways to make select the 4 appetizers. Put another way: there are 5 ways to not select a certain appetizer. We will use this value later, so let x = 5.


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Repeat for the main courses
n = 4 main courses overall, r = 3 selections allowed
*[Tex \Large _n C _r = \frac{n!}{r!*(n-r)!}]


*[Tex \Large _4 C _3 = \frac{4!}{3!*(4-3)!}]


*[Tex \Large _4 C _3 = \frac{4!}{3!*1!}]


*[Tex \Large _4 C _3 = \frac{4*3*2*1}{3*2*1*1}]


*[Tex \Large _4 C _3 = \frac{24}{6}]


*[Tex \Large _4 C _3 = 4]


Similar as before, we have four ways to not pick a main course (equivalent to picking the 3 other main courses). We'll use this value later, so let y = 4.


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Repeat for the desserts. We have n = 12 and r = 10 this time. Keep in mind that {{{n >= r}}} is always true.


*[Tex \Large _n C _r = \frac{n!}{r!*(n-r)!}]


*[Tex \Large _{12} C _{10} = \frac{12!}{10!*(12-10)!}]


*[Tex \Large _{12} C _{10} = \frac{12!}{10!*2!}]


*[Tex \Large _{12} C _{10} = \frac{12*11*10!}{10!*2*1}]


*[Tex \Large _{12} C _{10} = \frac{12*11}{2*1}] Note how the 10! terms cancel


*[Tex \Large _{12} C _{10} = \frac{132}{2}] 


*[Tex \Large _{12} C _{10} = 66] 


There are 66 ways to select ten desserts from a pool of twelve. Let z = 66.


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The last step is to multiply the values of x, y, and z found earlier.


x*y*z = 5*4*66 = 1320


Answer: <font color=red size=4>1320</font>
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