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# Need help with a math problem.

Discussion in 'The Okie Corral' started by alexanderg23, Mar 30, 2012.

1. ### alexanderg23

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I need some help, if anyone is good with math.
He's the question.

We're trying to figure out how many different combinations are possible for a sandwich meal at our restaurants.

Here's the info.

We have a sandwich plate that comes with 3 sandwiches, the chicken on the sandwich can come either come grilled or fried, and you can have 2 sauces out of 13 possible sauces on the sandwiches. you can mix and match the 3 sandwiches however you want grilled or fried and the sauces can be different on all three.

Whoever comes up with the right answer gets a big THANK YOU and a free meal if you ever come to one of our restaurants. Slim-chickens.com

2. ### 686OwnerNRA Life Member

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can you get 1 sauce only and is that the same as getting the same sauce twice? Is the bun optional? I'll assume the chicken is mandatory.

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4. ### 686OwnerNRA Life Member

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quick answer is 78 different combinations on the sauces (when you HAVE to get 2) times 3 sandwiches times 2 kinds of chicken

468

5. ### blackjack

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Not claiming math whiz status in any way but it seems the answer should be 3 * 2 * 13

3 = the number of sandwich occurences on the plate

2 = grilled or fried

13 = potential sauce choices, being limited to 2 is a distractor in this problem.

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7. ### alexanderg23

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I know without sauces there are 4 different combinations on how to get the sandwiches.

3 fried
3 grilled
1F/2G
2G/1F

it would be 6 but 1F/2G is that same as 2G/1F

8. ### alexanderg23

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is it 364 which is 91X4?

9. ### 686OwnerNRA Life Member

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Good call. So 4 times the number of sauce combinations, which was 91 above (assuming none, one, and extra of the same is allowed).

364 doing the multiplication above?

10. ### JimmyN

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It should be:
3 F
3 G
2F / 1G
2G / 1F

11. ### alexanderg23

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I thought there would be a lot more...

12. ### Z71bill

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I think factorials may be involved -

I hate factorials!

13. ### Lonestar 48Silver Member

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What's wrong with factories? We need factories to make the stupid sandwiches up in Arkansas.

14. ### MaboWant a beer?

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Ok here is what I came up with... I'm not a math genius by any means so take it with a grain of salt.

Ok like you surmised there are four possible sandwich combinations, lets call A fried and B grilled:

List has 4 entries.
{A,A,A} {A,A,B} {A,B,B} {B,B,B}

Now we have 13 different sauces, lets call them "a" through "m". We also have the possibility of no sauce "n" for a total of 14 different options. You said we could only pick two sauces, so here is what we come up with:

List has 105 entries.
{a,a} {a,b} {a,c} {a,d} {a,e} {a,f} {a,g} {a,h} {a,i} {a,j} {a,k} {a,l} {a,m} {a,n} {b,b} {b,c} {b,d} {b,e} {b,f} {b,g} {b,h} {b,i} {b,j} {b,k} {b,l} {b,m} {b,n} {c,c} {c,d} {c,e} {c,f} {c,g} {c,h} {c,i} {c,j} {c,k} {c,l} {c,m} {c,n} {d,d} {d,e} {d,f} {d,g} {d,h} {d,i} {d,j} {d,k} {d,l} {d,m} {d,n} {e,e} {e,f} {e,g} {e,h} {e,i} {e,j} {e,k} {e,l} {e,m} {e,n} {f,f} {f,g} {f,h} {f,i} {f,j} {f,k} {f,l} {f,m} {f,n} {g,g} {g,h} {g,i} {g,j} {g,k} {g,l} {g,m} {g,n} {h,h} {h,i} {h,j} {h,k} {h,l} {h,m} {h,n} {i,i} {i,j} {i,k} {i,l} {i,m} {i,n} {j,j} {j,k} {j,l} {j,m} {j,n} {k,k} {k,l} {k,m} {k,n} {l,l} {l,m} {l,n} {m,m} {m,n} {n,n}

So lets pretend that I got sandwich choice {A,A,A}, with sauce choice {b,c}. I could make the following combinations:

A/b A/b A/b
A/b A/b A/c
A/b A/c A/c
A/c A/c A/c

But it doesn't make sense for me to put sauce "c" or "b" on all three of my sandwiches otherwise I would have just gotten sauce combination {b,b} or {c,c}.

So if you get two different sauces there are two ways you would probably sauce your food:

A/b A/b A/c
A/b A/c A/c

If you get two of the same sauce there is only one way to sauce your food:

A/b A/b A/b

Ok so there are 14 sauce combinations that result in just one way to sauce your food, {a,a}, {b,b}, {c,c}... etc. While there are 91 double sauce combinations that allow you two ways to sauce your food. We know there are four possible sandwich combinations so here is what I come up with...

[(4x14)x1] + [(4x91)x2] = 56+728 = 784

4 sandwich combinations times 14 sauces yielding 1 way to sauce my food + 4 sandwich combinations times 91 double sauces yielding 2 ways to sauce my food = 784 choices

I'm probably wrong but at least it is a starting point and someone can see where I screwed up and hopefully we can figure this out.

Last edited: Mar 30, 2012
15. ### MaboWant a beer?

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Bumping to see if someone has a better answer.

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13*13=169
169*6=1014

17. ### DoubleWide

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So I tried to do math on a Saturday morning and that sucked.

Does this take into account that if you got two (or three) of the same sandwich, it's not really a new combination? like how fried and grilled don't make 6. It probably doesn't matter for marketing.

18. ### dherlocX-Nuc

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You are making me hungry.

19. ### glockaviator

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The question is ambiguous. Several ways of interpreting the OP's post.

I'll say this. You can have 3 sandwiches. I read that to mean all three the same, two the same and one other, all three different. And from three different meats (my interpretation). That is 3 ways to choose the first sandwich, 3 ways to choose the second sandwich and 3 ways to chose the third sandwich (again since you can have 3 of the same sandwich and 2 of the same sandwich).

That is 3 cubed, which is 27 ways to choose from 3 sandwiches (say chicken, beef or fish), allowing repeating.

You begin to see how complicated this can get.

Now if there are 2 ways of making each sandwich different, you now have 6 different sandwiches. That changes it to 6 cubed which is 216.

Not sure how to handle the 2 out of 13 though. With enough thought there is a way. If there are 13 ways of making each sandwich different, thats 6+13 = 19 cubed which is 6859. But no, that is not it because it is 2 out of 13.

If we ask how many ways are there to make a sandwich haveing exactly 2 sauces picking from 13 and allowing repeating (2 sauces the same), that is answerable. There are 13 ways to choose the first sauce and 13 ways to choose the second sauce. That is 13 x 13 =169.

So now we have 6 + 169 = 175 cubed which is 5,359,375. But I've probably made a mistake somehow. It's a complicated problem.

Last edited: Mar 31, 2012

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