Thanks for pointing out that Error. I am ashamed.

I wasn’t satisfied with that answer though.

This price seems a little strange for a buffet unless drinks were a seperate price and each person had a drink with a slightly different price. The resulting number would then include the average drink price. If two people had a large drink at $2.19 a piece and the 3rd had water, Each meal would have cost $12.99.

$12.99 per person seems more reasonable for a buffet lunch in Disneyworld.

That’s because I treated the identities of the three eaters as significant. I assumed-slash-stipulated that we would count the situation in which Dad pays $15.40 and Professor pays $16.10 as different from the situation in which Dad pays $16.10 and Professor pays $15.40. If you decide that only the prices, rather than who paid what, are significant, than yes, different answer.

Calling each solution a “combination,” as Kate did, might suggest that order is irrelevant, because one would say “permutation” if order is significant. I’d argue, however, that this isn’t a problem of combinations and permutations at all: we’re actually counting “partitions.”

Kate, I still do not see the relevance of your Chihuahua comment. To be honest, when I first read it, I just thought it had something to do with the language barrier. While we are on that topic, will I be invited to your home to celebrate the holidays this year? Just askin’.

Oh, c’mon. If I wrote “Elizabeth ate a fillet mignon,� wouldn’t your first reaction be “aha! We can’t just divide the total bill into three equal parts!�? So, why is the substitution of a cow for a dog made it so confusing? Are you saying it never occurred to you that the best Chihuahua is grilled Chihuahua?

I thought we already had everything set up for the holidays, no? You babysit our kid; we are out skiing. We already ordered a full set of Saved by the Bell DVDs, as you asked. Has anything changed?

BTW, did your server gets 18% of the buffet price pre or post discount? Inquiring minds want to know, as this is one difference between Rob’s and Urijah’s answers. But either way, neither price seems Disneyesque. Perhaps pricing wasn’t considered (an omission clearly “not in good faith”). One also wonders how much gratuity would be added if you’d actually got waiter service …

Just so that we are clear, this collective body has determined that the price per person of the Fort Wilderness buffet is between roughly $14 and $9 million. Good for us.

Let me say two things: 1. Each person’s buffet price was the same, and the 20% off applied to each person’s meal. 2. I am pretty sure I have the total lunch price wrong. I’ll have to ask my sister to check her Mastercard statement. I will report back.

God bless you all – I have not laughed this hard in a while. And, Kate, I still do not see the relevance of your Chihuahua comment. To be honest, when I first read it, I just thought it had something to do with the language barrier. While we are on that topic, will I be invited to your home to celebrate the holidays this year? Just askin’.

P.S. Antony Page, good to have you here.

James: my math “skills” are archaic, but I don’t see how your formula takes into account duplicate combinations. In other words, if the three meals are priced at x, y, z, we have only one combination of prices regardless of who consumes which meal. (i.e. Dad Nowicki eating the meal priced at x in x,y,z; the y in y,x,z; or the z in z,x,y; are all only one combination of prices.)

There’s a total of $44.34 to be divided among three meal prices. That’s 4434 cents to be partitioned among three labelled meals. The condition that each meal costs at least one cent means that this problem is equivalent to partitioning out 4431 cents among three labelled, possibly empty, meals. This problem, in turn, is equivalent to picking (in order) two integers whose sum is less than or equal to 4431 (the price of the third meal being whatever is left). We can write this expression as:
\Sum_{i=0}^{4431} \Sum_{j=0}^{4431-i} 1.

The outer sum corresponds to picking the price of Dad Nowicki’s meal. The inner sum corresponds to picking the price of Professor Nowicki’s meal, after taking out the price of Dad Nowicki’s meal. The 1 corresponds to the single unique price left for Other Professor Nowicki’s meal once the prices for the first two meals have been set. This simplifies to

\Sum_{i=0}^{4431} (4431 – i + 1)

because the inner sum is equal to the number of terms in the sum. (n times 1 is n.) We can now add the 1 to the 4431 to get 4432:

\Sum_{i=0}^{4431} (4432 – i)

Splitting the sum over the two inner components gives:

(\Sum_{i=0}^{4431} 4432) – (\Sum_{i=0}^{4432} i)

The first term here is easy; it’s just 4432 added up 4432 times (since the index i starts at 0):

(4432 * 4432) – (\Sum_{i=0}^4432 i)

The second term is only slightly more complicated. It’s a triangular number. Using the formula for the sum of the first n integers — (n * (n-1)/2) — we can rewrite our equation as:

4432^2 – ((4432)(4431)/2)

Doing out the multiplications, we get:

19,642,624 – (19,638,192/2)

Performing the division by 2 gives:

19,642,624 – 9,819,096

Which yields a final answer of:

9,823,528

Next question: contrary to James’ comment above, the number of possible solutions isn’t quite infinite, so long as meal prices involve only whole cents. Hence the question: how many combinations of meal prices are possible, assuming that no meal costs less than 1 cent, each meal costs something, and prices involve only whole cents?