Comments on: A Summer Hamburger Challenege

By: amy

amy — Fri, 30 Jun 2006 15:27:54 +0000

good lord. forget the possible hambuger combinations and hand me a chocolate malt shake…its one of the best

By: Cant_picture_it

Cant_picture_it — Fri, 30 Jun 2006 00:20:33 +0000

Wow, ime gettin’ edjamacated.

By: Paulo

Paulo — Thu, 29 Jun 2006 23:27:05 +0000

There’s still a Burrito Brothers in the Union Station basement food court. I stop by there at least once a week for a super spinach.

By: LD

LD — Thu, 29 Jun 2006 20:59:23 +0000

The only thing more riveting than this “Mathletes” smackdown was the Great Spelling Bee Conspiracy of 2006.

I’m going to McDonalds.

By: Don

Don — Thu, 29 Jun 2006 19:44:23 +0000

Crap-o, Al’s right – I did a quick check in my head before I posted that and didn’t even notice that I failed to count combos with all items and no items.

My theory of algorithms prof would be ashamed of me.

And NC – I’d say how completely I forgot my permutation/combination formulas proves I’m a comp sci turned writer type, not a math geek.

By: al

al — Thu, 29 Jun 2006 18:34:03 +0000

Don, not quite. 18! would be the number of 18-member permutations you could make from a set of 18 total members.

You’re quite right that we want combinations, not permutations, and the general formula for that is n!/((n-k)!*k!), where k is the number of elements in the combination, and n is the total number of elements from which to choose.

Wrinkle is that, while our n=18 (or whatever), we don’t have a fixed k: it can range from 0 to n, since we can have none of the toppings, some of them, or all of them. We could sum up this equation for each value of k, or we could just do it the way I did it in the first place, which is equivalent.

By: Number Challeneged

Number Challeneged — Thu, 29 Jun 2006 18:28:22 +0000

math geek fight! i thought this was an english major turned lawyer town.

By: Al

Al — Thu, 29 Jun 2006 18:22:08 +0000

Um, no. I don’t know where you’re coming up with your method of calculation, but it’s just plain wrong. Might be a little counterintuitive, but the 2^18th is quite right.

Can’t wrap your mind around it in the abstract? Try this illustration. Suppose the only things you could put on a burger were ketchup, mustard, and/or cheese (K, M, and C) and you could only cook the burger rare or well-done (R or W). That’s 3 toppings, 2 ways to cook. By your formula, that’s 3×2 = 6 combos. By mine, it’s (2^3)*2 = 16. Which sounds right?

Still don’t believe me? Count ‘em: plain-R, plain-W, K-R, K-W, M-R, M-W, C-R, C-W, KM-R, KM-W, KC-R, KC-W, MC-R, MC-W, KMC-R, KMC-W.

Got it now?

By: wayan

wayan — Thu, 29 Jun 2006 18:20:25 +0000

And 18 is less than 1,000,001, the number (over 1 million) that we need to reach in order for the Johnny Rockets claim to be true.

Again, where is that bored and hungry lawyer?

By: Don

Don — Thu, 29 Jun 2006 18:16:27 +0000

You’re both wrong. Wayan, your numbers ignore that if you have toppings A, B and C you actually have more than 3 combinations – 6, in fact.

Al, you’re failing to recognize that the combinations are not order-dependent – this is a set. A burger with A & B is the same as a burger with B & A.

The number of topping combinations is actually 18!, 18 factorial, times 4. NOT 2 raised to the power of 18 times 4. Google calculates that as 6.40237371 × 10^15

And actually, you need to add 1 to that: no toppings.