Arby's and Combinatorics

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Arby's has this deal out now: 5 for $5.95. You can pick from 8 choices to fill 5 spots. The menu says "over 790 possible combinations" , so one can assume that the number of choices is between 790 and 800. What is the equation to figure out how many choices you have and what is the exact number of possible combinations?
 

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  • #2
1,425
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There are well over 800 possibilities.
 
  • #3
1,074
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There are well over 800 possibilities.
How much is well over 800? And how do you figure that out?
 
  • #4
1,425
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Anyway to question is posed the number of possibilities is over 800; if one selection is independent of the others, there are 8^5 possibilities and if it is, there are 8*7*6*5*4 possibilities.
 
  • #5
Integral
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http://en.wikipedia.org/wiki/Combinatorics" [Broken] has the answer. In this case, order does not matter and and you can order the same thing more then once. (2 Arby sandwichs and 3 frys)

edit: I get 792
 
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  • #6
cristo
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Anyway to question is posed the number of possibilities is over 800; if one selection is independent of the others, there are 8^5 possibilities and if it is, there are 8*7*6*5*4 possibilities.
The answer is not "way more than 790," but is actually 792. Check the "combination with repetition" section of this page of wikipedia.
 
  • #7
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Anyway to question is posed the number of possibilities is over 800; if one selection is independent of the others, there are 8^5 possibilities and if it is, there are 8*7*6*5*4 possibilities.
All selections are independant, but 8^5 doesn't take into account any duplications of choices this is what I thought at first as well and realized it made no sense.
 
  • #8
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Oops, sorry, I just realized.
 
  • #9
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Another related question..

Say I have 3 items A,B,C and I wish to select 3 at a time (order does not matter and an object can be chosen more than once)...

using the forumla for Combination with repetition, I should have (3+3-1)!/3!*2!=10 combinations

If I list all the possible combinations, I have:

AAA BBB CCC
AAB BBC
AAC BCC
ABB
ACC

But this is only 9.....what combination am I missing?
 
  • #10
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sorry.. didn't look at the most obvious choice.. ABC!!!
 
  • #11
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lol wow i was just about to post a topic about this and i searched google first to see if anyone had.

i was curious as to why they said over 790 choices, as that would seem to imply it's between 790 and 800 (or else they would have said over 800 choices).

there are 8 choices, and 5 slots to put them in. that should be 8^5 if i remember from discrete math. 8^5 is a lot larger than 790 though.
 
  • #12
cristo
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there are 8 choices, and 5 slots to put them in. that should be 8^5 if i remember from discrete math. 8^5 is a lot larger than 790 though.
Read post #5 or #6.
 
  • #13
cepheid
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there are 8 choices, and 5 slots to put them in. that should be 8^5 if i remember from discrete math. 8^5 is a lot larger than 790 though.
Er, yeah, because the order of the elements doesn't matter. You're just choosing, not permuting. If you order two medium drinks, two curly fries and a roast beef sandwich, that's the same as ordering a roast beef sandwich, an order of curly fries, a medium drink, another order of curly fries, and a medium drink. You've put in the same food order, you just haven't requested the items in the same sequence. The two examples are not distinct combinations of five items.

What Cristo said, if you want to see the mathematics.

By the way, here in Canada, you still pay $6, but you only get to choose four items! I don't understand why that is. The menu says 330 possible combinations, which leads me to believe that there are still 8 items to choose from (if you use the formula provided in the Wikipedia link, it works out exactly).

That leads me to my next question. Does anyone know how to "derive" or at least intuitively explain this formula for combination with repetition? Because, all three of the other combinatorial formulas make perfect sense to me (and if you don't believe me, I'd be happy to explain why I think they make sense intuitively). But combination with repetitition...looking at the formula, I can't seem to figure out how one might arrive at it.
 

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